acids ph buffers a2 chemistry

Acids ph buffers A2 chemistry

Using the post it notes write down the name of the species and whether each species round the

room is an acid, base or alkali

H3O+

pka = -1.7

H2O

pka = 15.7

HI

pka = -10

NH4+

pka = 9.2

Al2O3

Na2O

SO2

Ammonia

pka = 36

Complete the equations match up

pka = 36

Loop game

0.8L of 1M ethanoic acid reacted was titrated with 0.2L 1M NaOH. Calculate the pH of the

solution. Ka = 1.76 x 10-5

Acid moles [HA] H+ Base moles

Initial

change

Equilibrium

Step 1. Work out moles of HA at start

Step 2. Work out moles of base at start

Step 3. Work out moles of HA that reacted with base – this is moles H+

Step 4. Minus moles of neutralised HA from initial acid to get the moles of HAKa x [HA] = [H+]2Step 5. Use equilibrium concentrations and put them in this equation

Step 6. use pH = -log [H+]

0.25L of 1M ethanoic acid reacted was titrated with 0.25L 1M Ca(OH)2. Calculate the pH of the

solution. Ka = 1.76 x 10-5

Acid moles [HA] H+ Base moles

Initial

change

Equilibrium

Step 1. Work out moles of HA at start

Step 2. Work out moles of base at start

Step 3. Work out moles of HA that reacted with base – this is moles H+

Step 4. Minus moles of neutralised HA from initial acid to get the moles of HAKa x [HA] = [H+]2Step 5. Use equilibrium concentrations and put them in this equation

Step 6. use pH = -log [H+]

0.3L of 1M Methanoic acid reacted was titrated with 0.7L 1M Ca(OH)2. Calculate the pH of the

solution. pKa = 3.77

Acid moles [HA] H+ Base moles

Initial

change

Equilibrium

Step 1. Work out moles of HA at start

Step 2. Work out moles of base at start

Step 3. Work out moles of HA that reacted with base – this is moles H+

Step 4. Minus moles of neutralised HA from initial acid to get the moles of HAKa x [HA] = [H+]2Step 5. Use equilibrium concentrations and put them in this equation

Step 6. use pH = -log [H+]

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