Question

You wish to titrate a 0.181 M solution of HCN (pKa = 9.30) with a 0.183 M solution of NaOH.

Part #1:

If you begin with 42.3 mL of HCN solution, how many mL of the NaOH solution should be added to reach the equivalence point?

Part #2:

What is the pH at the equivalence point?

Answer 1

1)

mol of acid = mol of base

then

mol of acid = MV = 0.181*42.3 = 7.6563 mmol of acid

then

M = mmol/mL

mL = mmol/M = 7.6563/(0.183) = 41.83 ml of base needed

2)

pH in equivalence

CN- + H2O <--> HCN + +OH-

Kb = [HCN][OH-]/[CN-]

Kb = (kw/Ka) = (10^-14)/(10^-9.3) = 0.00001995262

0.00001995262 = (x*x)/(M-x)

M = mmol of CN- / total V

Total V = V1+V2 = 42.3 + 41.83 = 84.13

then

M = 7.6563/(84.13) = 0.09100

0.00001995262 = (x*x)/(0.09100)

x = sqrt(0.00001995262*0.09100) = 0.00134

OH- = .00134

pOH = -log(.00134) = 2.87

ph = 14-2.87 = 11.13