# 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.

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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