Question

Acid-Base Titration: A 0.15 M solution of NaOH is used to titrate 200.0 mL of 0.15 M HCN. What is the pH at the equivalence point? Ka = 4.9 x 10-10.

Please include steps/work so I can see and understand how this problem is solved.

Answer 1

one mole NaOH neutralize the one mole HCN.

so

at neutralization point,

volume of NaOH needed = 200.0 mL * 0.15 M / 0.15 M = 200.0 ml

thus

at neutralization point, volume of solution mixture = (200 + 200) = 400 ml

and

concentration of salt =  200.0 mL * 0.15 M / 400 ml = 0.075 M

Ka = 4.9 x 10-10

pKa = - log (4.9 x 10-10) = 9.31

salt hydrolysis occurs.

so

Using

pH = 1 / 2 * [pKw + pKa + log C]

or

pH = 0.5 * [14 + 9.31 + log (0.075)]

or

pH = 11.1 (answer)

pH at the equivalence point = 11.1