1. In the titration of 25.00mL
1. In the titration of 25.00mL of 0.10M HCN using 0.10M NaOH asthe titrant, what is the pH after the addition of 12.50mL ofNaOH?
2. In the titration of 25.00mL of 0.10M HCN using 0.10M NaOH asthe titrant, what is the pH after the addition of 25.00mL ofNaOH?
3. What is the pH of 0.050M NaCN? Ka for HCN is 4.9×10-10.
Answer:
1)
nuetrilization reaction between HCN and NaOH is
NaOH + HCN 
NaCN +H2O
no. of mole = molarity 
volume ofsolution in liter
no. of mole of HCN = 0.10 0.025 = 0.0025mole
no. of mole of NaOH = 0.10 0.01250 =0.00125 mole
0.00125 mole of NaOH react with 0.00125 mole of HCN to form saltand water
mole of HCN remain in solution = 0.0025 – 0.00125 = 0.00125mole
total volume of solution = 25 + 12.50 = 37.50 ml = 0.03750liiter
molarity = no. of mole / volume of solution in liter
molarity of HCN = 0.00125 / 0.03750 = 0.0333 M
HCN dissociate as
HCN + H2O 
H3O+ + CN–
Ka = [H3O+][CN–] / [HCN]
but [H3O+] = [CN–] = x
then Ka = [x] [x] / [HCN]
[x]2 = Ka [HCN] = 4.910-10 0.0333 = 1.631710-11
[x] = 4.039 10-6 = [H3O+]= [CN–] = x
pH = -log[H3O+] = -log(4.039 10-6) = 5.39
2)
nuetrilization reaction between HCN and NaOH is
NaOH + HCN NaCN +H2O
no. of mole = molarity volume ofsolution in liter
no. of mole of HCN = 0.10 0.025 = 0.0025mole
no. of mole of NaOH = 0.10 0.025 = 0.0025mole
0.0025 mole of NaOH react with 0.0025 mole of HCN to form saltand water
mole of HCN remain in solution = 0.00 mole
pH = 7
3)
HCN dissociate as
HCN + H2O H3O+ + CN–
Ka = [H3O+][CN–] / [HCN]
but [H3O+] = [CN–] = x
then Ka = [x] [x] / [HCN]
[x]2 = Ka [HCN] = 4.910-10 0.050 = 2.4510-11
[x] = 4.95 10-6 = [H3O+]= [CN–] = x
pH = -log[H3O+] = -log(4.95 10-6) = 5.30
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