here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate...
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![Page 1: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/1.jpg)
Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture.
Strong Acid–Strong Base
Mixture CalculationsExample 2
![Page 2: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/2.jpg)
We’re given that 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
![Page 3: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/3.jpg)
And we’re asked to determine the pH of the final mixture.
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
![Page 4: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/4.jpg)
Just a few words about sulphuric acid, H2SO4.
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
![Page 5: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/5.jpg)
H2SO4.
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
H2SO4
![Page 6: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/6.jpg)
is a Diprotic Acid
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
H2SO4 is a Diprotic Acid
![Page 7: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/7.jpg)
Which means it has 2 protons it can lose.
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
H2SO4 is a Diprotic AcidIt has 2 protons it can lose.
![Page 8: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/8.jpg)
As soon as H2SO4 is added to water, it ionizes completely to lose its first proton:
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
When H2SO4 is added to water, it ionizes completely to lose its first proton:
![Page 9: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/9.jpg)
100% of the H2SO4 molecules lose one proton (click) to form hydronium and hydrogen sulphate ions.
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
When H2SO4 is added to water, it ionizes completely to lose its first proton:
2 4 2 3 4H SO H O H O HSO 100% ionization
H+
![Page 10: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/10.jpg)
But when its just in water, the second proton does not come off as easily. This proton comes off when HSO4 minus ionizes. But HSO4- is a weak acid so its ionization in water is very limited.
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
When H2SO4 is added to water, it ionizes completely to lose its first proton:
2 4 2 3 4H SO H O H O HSO 100% ionization
But its second proton does not come off as easily in water:
24 2 3 4HSO H O H O SO 100% ionization
EquilibriumWeak Acid
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However, when H2SO4 is mixed with the STRONG BASE KOH, this is a totally different situation.
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
When H2SO4 is mixed with thestrong base KOH, this is a totally different situation.
![Page 12: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/12.jpg)
When an H2SO4 molecule enters water, it loses one proton (click) to water, to form a hydronium ion (H3O+) and a hydrogen sulphate ion (HSO4 minus).
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
2 2 434 HH S O H OO OH S H+
![Page 13: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/13.jpg)
Models of these are shown here. Take a moment to check the atoms and the charges and see how the formulas relate to the structural models.
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
OH
H
H +
O
O H
SO
O–
2 2 434 HH S O H OO OH S H+
![Page 14: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/14.jpg)
When the strong base KOH dissociates in water it forms K+ and OH minus ions. Here we doubled everything in the equation.
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
2KOH 2K+ + 2OH–
OH
H
H +
O
O H
SO
O–
![Page 15: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/15.jpg)
we show models of the two hydroxide ions from the KOH
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
OH
HO
–
–
2KOH 2K+ + 2OH–
OH
H
H +
O
O H
SO
O–
![Page 16: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/16.jpg)
One of the hydroxide ions collides with the hydronium ion
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
HO
OH
–
–
OH
H
H +
O
O H
SO
O–
![Page 17: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/17.jpg)
and takes away a proton, to form 2 water molecules
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
HO–
OH
H
H OH
O
O H
SO
O–
![Page 18: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/18.jpg)
The other hydroxide ion collides with the hydrogen sulphate ion (click)
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
HO
H OH
–
OH
H
O
O H
SO
O–
![Page 19: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/19.jpg)
And takes a proton from it to form a water molecule and a suphate ion
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
O
OS
O
O–
H OH
HO
H
–
OH
H
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the sulphate ion has the formula SO4 2 minus.
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
H OH
HO
H
O
OS
O
O–
–
Sulphate SO4
2–
OH
H
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So, in an indirect way, 2 hydroxide ions are able to remove both protons from a molecule of H2SO4. We‘ll show this with equations.
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
Two OH– ions were able to remove both protons from H2SO4.
![Page 22: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/22.jpg)
As soon as H2SO4 is added to water it ionizes completely to form a hydronium ion and a hydrogen sulphate ion. We’ll call this Step 1
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
Two OH– ions were able to remove both protons from H2SO4.
2 4 2 3 4H SO H O H O HSO Step 1
![Page 23: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/23.jpg)
When we add a strong base, one OH minus ion neutralizes the hydronium ion to form 2 water molecules. We’ll call this step 2
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
Two OH– ions were able to remove both protons from H2SO4.
2 2 434 HH SO O H OOH S
OH+
22H O
Step 2
![Page 24: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/24.jpg)
And the other OH minus ion reacts with hydrogen sulphate to form water and a sulphate ion. We’ll call this Step 3.
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
Two OH– ions were able to remove both protons from H2SO4.
42 4 2 3H SO H O H HO SO
OH+
22H O
OH+
22 4H O SO
Step 3
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Even though we know these 3 steps occur when we add H2SO4 to water and then add a strong base,
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
Even though we know that these steps occur…
1
2 3
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We can represent the process with a net overall equation: H2SO4 plus 2 OH minus form 2H2O plus SO4 2minus.
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
Even though we know that these steps occur…
22 42 4H SO 2OH 2H O SO
We can represent the process with a net overall equation.
1
2 3
![Page 27: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/27.jpg)
so in the overall net reaction, we see that each H2SO4 (click) donates 2 protons or H+ ions to the hydroxide ions.
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
422
2 4SO 2OHH 2H O SO 2 H+
Each H2SO4 donates
2 protons to the OH– ions
![Page 28: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/28.jpg)
From this, we can write the conversion factor stating there are 2 moles of H+ per 1 mole of H2SO4. We can use this conversion factor in any calculation where H2SO4 reacts with a strong base.
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
422
2 4SO 2OHH 2H O SO
2 H+
Each H2SO4 donates
2 protons to the OH– ions 2 4
2 mol H
1 mol H SO
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Now we’ll do the calculations for this problem. We’ll begin by calculating the initial moles of H+ added.
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
2 4
2 4
0.150mol H SO 2momol
l H0.125L 0.0375mol H
1L 1mol H SOHinitial
0.200mol KOH 1mol OHmol OH 0.150L 0.0300mol OH
1L 1mol KOHinitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 0.150L 0.275L
3pH log H O log 0.0273 1.56 2 4
2 mol H
1 mol H SO
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Its equal to 0.150 moles of H2SO4 per Litre…
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
2
2 4
4 2mol Hmol H 0.125L 0.0375mol H
1
0.150
mol H SO
mol H SO
1Linitial
0.200mol KOH 1mol OHmol OH 0.150L 0.0300mol OH
1L 1mol KOHinitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 0.150L 0.275L
3pH log H O log 0.0273 1.56 2 4
2 mol H
1 mol H SO
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Times 2 moles of H+ to 1 mole of H2SO4…
2
2 4
4 2mol H
1mo
0.150mol H SOmol H 0.125L 0.0375mol H
1 l H SL Oinitial
0.200mol KOH 1mol OHmol OH 0.150L 0.0300mol OH
1L 1mol KOHinitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 0.150L 0.275L
3pH log H O log 0.0273 1.56 2 4
2 mol H
1 mol H SO
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
![Page 32: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/32.jpg)
Times 0.125 L
2 4
2 4
0.150mol H SO 2mol Hmol H 0.0375mol0 H
1L 1mol SO5L
H.12initial
0.200mol KOH 1mol OHmol OH 0.150L 0.0300mol OH
1L 1mol KOHinitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 0.150L 0.275L
3pH log H O log 0.0273 1.56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
![Page 33: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/33.jpg)
Which comes to 0.0375 moles of H+. Notice moles of H2SO4 and Litres cancel out.
2 4
2 4
0.150mol H SO 2mol Hmol 0.0375molH 0.125L
1L 1mol H SOHinitial
0.200mol KOH 1mol OHmol OH 0.150L 0.0300mol OH
1L 1mol KOHinitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 0.150L 0.275L
3pH log H O log 0.0273 1.56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
![Page 34: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/34.jpg)
In order to preserve 3 significant figures (the lowest number of significant figures in the given data)…
2 4
2 4
0.150mol H SO 2mol Hmol H 0.125L 0.0 mol37 H
1L 1mol H SO5initial
0.200mol KOH 1mol OHmol OH 0.150L 0.0300mol OH
1L 1mol KOHinitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 0.150L 0.275L
3pH log H O log 0.0273 1.56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
3 significant
figures
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The answer to this must be expressed to 4 decimal places.
2 4
2 4
0.150mol H SO 2mol Hmol H 0.125L 0. mol037 H
1L 1mol H S5
Oinitial
0.200mol KOH 1mol OHmol OH 0.150L 0.0300mol OH
1L 1mol KOHinitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 0.150L 0.275L
3pH log H O log 0.0273 1.56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
4 decimal places
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Now we’ll calculate the initial moles of OH minus added.
2 4
2 4
0.150mol H SO 2mol Hmol H 0.125L 0.0375mol H
1L 1mol H SOinitial
0.200mol KOH 1mol OH0.150L 0.0300mol OH
1L 1molmol
KH
OHO initial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 0.150L 0.275L
3pH log H O log 0.0273 1.56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
![Page 37: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/37.jpg)
It is equal to 0.200 moles of KOH per L
2 4
2 4
0.150mol H SO 2mol Hmol H 0.125L 0.0375mol H
1L 1mol H SOinitial
1mol OHmol OH 0.150L 0
0.2.0300mol OH
1mol KO
00mo
H
l KOH
1Linitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 0.150L 0.275L
3pH log H O log 0.0273 1.56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
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Times 1 mole of OH minus to 1 mole of KOH
2 4
2 4
0.150mol H SO 2mol Hmol H 0.125L 0.0375mol H
1L 1mol H SOinitial
1mol OH0.200mol KOHmo
1mol Kl OH 0.150L 0.0300mol OH
OH1Linitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 0.150L 0.275L
3pH log H O log 0.0273 1.56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
![Page 39: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/39.jpg)
Times 0.150 L
2 4
2 4
0.150mol H SO 2mol Hmol H 0.125L 0.0375mol H
1L 1mol H SOinitial
0.200mol KOH 1mol OHmol OH 0.0300mol OH
1L 1mol0.150L
KOHinitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 0.150L 0.275L
3pH log H O log 0.0273 1.56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
![Page 40: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/40.jpg)
Which comes out to 0.0300 moles of OH minus. You can see that moles of KOH and Litres both cancel. Notice we also have 3 significant figures and 4 decimal places in this answer.
2 4
2 4
0.150mol H SO 2mol Hmol H 0.125L 0.0375mol H
1L 1mol H SOinitial
0.200mol KOH 1mol OHmol OH 0.150L
1L 1mol0.0300mol
KOHOHinitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 0.150L 0.275L
3pH log H O log 0.0273 1.56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
![Page 41: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/41.jpg)
Now we compare the initial moles of H+ and OH minus. We see that 0.0375, the moles of H+, is greater than 0.0300, the moles of OH minus
2 4
2 4
0.150mol H SO 2mol Hmol 0.0375molH 0.125L
1L 1mol H SOHinitial
0.200mol KOH 1mol OHmol OH 0.150L
1L 1mol0.0300mol
KOHOHinitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 0.150L 0.275L
3pH log H O log 0.0273 1.56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
![Page 42: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/42.jpg)
So the H+ is in excess and the OH minus is the limiting reagent.
2 4
2 4
0.150mol H SO 2mol Hmol 0.0375molH 0.125L
1L 1mol H SOHinitial
0.200mol KOH 1mol OHmol OH 0.150L
1L 1mol0.0300mol
KOHOHinitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 0.150L 0.275L
3pH log H O log 0.0273 1.56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
In Excess
Limiting Reagent
![Page 43: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/43.jpg)
We calculate the excess moles of H+…
2 4
2 4
0.150mol H SO 2mol Hmol H 0.125L 0.0375mol H
1L 1mol H SOinitial
0.200mol KOH 1mol OHmol OH 0.150L 0.0300mol OH
1L 1mol KOHinitial
+ 0.0375mol H 0.0300mol OHExce 0.0075mol Hssmol H
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 0.150L 0.275L
3pH log H O log 0.0273 1.56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
![Page 44: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/44.jpg)
By taking 0.0375 moles of H+
2 4
2 4
0.150mol H SO 2mol Hmol 0.0375molH 0.125L
1L 1mol H SOHinitial
+Excessmol H 0.0300mol OH0.0375mo 0.0075molH Hl
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 0.150L 0.275L
3pH log H O log 0.0273 1.56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
0.200mol KOH 1mol OHmol OH 0.150L 0.0300mol OH
1L 1mol KOHinitial
![Page 45: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/45.jpg)
and subtracting 0.0300 moles of OH minus.
2 4
2 4
0.150mol H SO 2mol Hmol H 0.125L 0.0375mol H
1L 1mol H SOinitial
+ 0.Exc 030essmol 0mol OHH 0.0375mol H 0.0075mol H
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 0.150L 0.275L
3pH log H O log 0.0273 1.56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
0.200mol KOH 1mol OHmol OH 0.150L
1L 1mol0.0300mol
KOHOHinitial
![Page 46: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/46.jpg)
To give us 0.0075 moles of H+ in excess
2 4
2 4
0.150mol H SO 2mol Hmol H 0.125L 0.0375mol H
1L 1mol H SOinitial
+Excessmol H 0.0375mol H 0.0300mol 0.0075mol HOH
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 0.150L 0.275L
3pH log H O log 0.0273 1.56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
0.200mol KOH 1mol OHmol OH 0.150L 0.0300mol OH
1L 1mol KOHinitial
![Page 47: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/47.jpg)
This answer, 0.0075, has 4 decimal places, because the numbers we subtracted both had 4 decimal places.
2 4
2 4
0.150mol H SO 2mol Hmol H 0.125L 0.0375mol H
1L 1mol H SOinitial
+Excessmol H 0. mol H 0. mol O0375 H 0. mol030 50 H007
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 0.150L 0.275L
3pH log H O log 0.0273 1.56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
0.200mol KOH 1mol OHmol OH 0.150L 0.0300mol OH
1L 1mol KOHinitial
4 decimal places
4 decimal places
4 decimal places
![Page 48: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/48.jpg)
But we can see that written this way, this number has only 2 significant figures, the 7 and the 5. Therefore the final answer to this problem cannot have more than 2 significant figures.
2 4
2 4
0.150mol H SO 2mol Hmol H 0.125L 0.0375mol H
1L 1mol H SOinitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.00 mol5 H7
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 0.150L 0.275L
3pH log H O log 0.0273 1.56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
0.200mol KOH 1mol OHmol OH 0.150L 0.0300mol OH
1L 1mol KOHinitial
2 significant figures
![Page 49: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/49.jpg)
The next step on the way to pH, is to find the hydronium ion concentration.
2 4
2 4
0.150mol H SO 2mol Hmol H 0.125L 0.0375mol H
1L 1mol H SOinitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
+ +
3
0.0075mol H 0.0075mol H0.0273M
0.125L 0.150L 0.275LH O H
3pH log H O log 0.0273 1.56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
0.200mol KOH 1mol OHmol OH 0.150L 0.0300mol OH
1L 1mol KOHinitial
![Page 50: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/50.jpg)
Which is equal to the concentration of H+. These are synonymous in chemistry dealing with aqueous solutions.
2 4
2 4
0.150mol H SO 2mol Hmol H 0.125L 0.0375mol H
1L 1mol H SOinitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
+ +
3
0.0075mol H 0.0075mol H0.0273M
0.125L 0.150L 0.275LH O H
3pH log H O log 0.0273 1.56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
0.200mol KOH 1mol OHmol OH 0.150L 0.0300mol OH
1L 1mol KOHinitial
![Page 51: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/51.jpg)
The concentration of H+ is equal to moles of H+ per Litre of solution. The moles of H+ is 0.0075 moles.
2 4
2 4
0.150mol H SO 2mol Hmol H 0.125L 0.0375mol H
1L 1mol H SOinitial
+Excessmol H 0.0375mol H 0.0300mol 0.0075mol HOH
+ +
3
0.0075mol HH O H 0.0273M
0.125L
0.0075mol H
0.150L 0.275L
3pH log H O log 0.0273 1.56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
0.200mol KOH 1mol OHmol OH 0.150L 0.0300mol OH
1L 1mol KOHinitial
![Page 52: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/52.jpg)
And the total volume of the mixture is 0.125 L of H2SO4 ….
2 4
2 4
0.150mol H SO 2mol Hmol H 0.0375mol0 H
1L 1mol SO5L
H.12initial
+ +
3 0.125L
0.0075mol H 0.0075mol HH O H 0.0273M
0.150L 0.275L
3pH log H O log 0.0273 1.56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
0.200mol KOH 1mol OHmol OH 0.150L 0.0300mol OH
1L 1mol KOHinitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
![Page 53: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/53.jpg)
Plus 0.150 L of KOH.
2 4
2 4
0.150mol H SO 2mol Hmol H 0.125L 0.0375mol H
1L 1mol H SOinitial
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 00.150L .275L
3pH log H O log 0.0273 1.56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
0.200mol KOH 1mol OHmol OH 0.0300mol OH
1L 1mol0.150L
KOHinitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
![Page 54: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/54.jpg)
So the concentration of H+ or H3O+ is 0.0075 moles over 0.275 L.
2 4
2 4
0.150mol H SO 2mol Hmol H 0.125L 0.0375mol H
1L 1mol H SOinitial
+ +
3
0.0075mol H0.0273M
0.125L 0.150L
0.0075mol HH O H
0.275L
3pH log H O log 0.0273 1.56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
0.200mol KOH 1mol OHmol OH 0.150L 0.0300mol OH
1L 1mol KOHinitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
![Page 55: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/55.jpg)
Which comes out to 0.0273 molar. We’ll carry one more significant figure than the 2 our final answer is limited to. We’ll round to 2 significant figures at the end.
2 4
2 4
0.150mol H SO 2mol Hmol H 0.125L 0.0375mol H
1L 1mol H SOinitial
+ +
3
0.0075mol H 0.0075mol H
0.125L 0.150LH O H 0.0273M
0.275L
3pH log H O log 0.0273 1.56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
0.200mol KOH 1mol OHmol OH 0.150L 0.0300mol OH
1L 1mol KOHinitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
![Page 56: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/56.jpg)
In the last step, we’ll find the pH of the mixture.
2 4
2 4
0.150mol H SO 2mol Hmol H 0.125L 0.0375mol H
1L 1mol H SOinitial
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 0.150L 0.275L
3log H O log 0.0273 1.56pH 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
0.200mol KOH 1mol OHmol OH 0.150L 0.0300mol OH
1L 1mol KOHinitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
![Page 57: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/57.jpg)
Remember, pH is defined as the negative log of the hydronium ion concentration.
2 4
2 4
0.150mol H SO 2mol Hmol H 0.125L 0.0375mol H
1L 1mol H SOinitial
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 0.150L 0.275L
3 log 0.0273 1.pH log H O 56 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
0.200mol KOH 1mol OHmol OH 0.150L 0.0300mol OH
1L 1mol KOHinitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
![Page 58: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/58.jpg)
Which is the negative log of 0.0273
2 4
2 4
0.150mol H SO 2mol Hmol H 0.125L 0.0375mol H
1L 1mol H SOinitial
++
3
0.0075mol H 0.0075mol HH M
0.12H O 0
5L 0.150L 0.275L.0273
3 log 0.02pH log H O 1.5673 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
0.200mol KOH 1mol OHmol OH 0.150L 0.0300mol OH
1L 1mol KOHinitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
![Page 59: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/59.jpg)
, which comes out to 1.56.
2 4
2 4
0.150mol H SO 2mol Hmol H 0.125L 0.0375mol H
1L 1mol H SOinitial
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 0.150L 0.275L
3log H O log 0.02pH 1.5673 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
0.200mol KOH 1mol OHmol OH 0.150L 0.0300mol OH
1L 1mol KOHinitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
2 significant figures
![Page 60: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/60.jpg)
In a pH, the digits to the right of the decimal are significant. So this answer has 2 significant figures.
2 4
2 4
0.150mol H SO 2mol Hmol H 0.125L 0.0375mol H
1L 1mol H SOinitial
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 0.150L 0.275L
3log H O log 0.02pH 5673 1. 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
0.200mol KOH 1mol OHmol OH 0.150L 0.0300mol OH
1L 1mol KOHinitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
2 significant figures
![Page 61: Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base](https://reader035.vdocuments.net/reader035/viewer/2022062315/5697bfdb1a28abf838cb05d9/html5/thumbnails/61.jpg)
Now we have answered the original question. The pH of the final mixture is 1.56. This low value means the solution is fairly acidic. This is reasonable because a strong acid is in excess in this case.
2 4
2 4
0.150mol H SO 2mol Hmol H 0.125L 0.0375mol H
1L 1mol H SOinitial
+ +
3
0.0075mol H 0.0075mol HH O H 0.0273M
0.125L 0.150L 0.275L
3log H O log 0.02pH 1.5673 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?
0.200mol KOH 1mol OHmol OH 0.150L 0.0300mol OH
1L 1mol KOHinitial
+Excessmol H 0.0375mol H 0.0300mol OH 0.0075mol H
pH of Final Mixture