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
H3
+
O
pka = -1.7
H2 O
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 HA
Step 5. Use equilibrium concentrations and put them in this equation  Ka x [HA] = [H+]2
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 HA
Step 5. Use equilibrium concentrations and put them in this equation  Ka x [HA] = [H+]2
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 HA
Step 5. Use equilibrium concentrations and put them in this equation  Ka x [HA] = [H+]2
Step 6. use pH = -log [H+]
Q1
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Q7
Q8
Q9
Q10