Question

A 50.00 mL sample of 0.350M hydrocyanic acid (HCN) is titrated with 0.250M cesium hydroxide (CsOH). What is the pH of the solution at the equivalence point? K_a(HCN) = 4.0x10^-10

HCN_(aq) + CsOH_(aq)<----> CsCN_(aq) + H₂O_(l)

Answer 1

The reaction between CsOH and HCN is 1:1 molar reaction

HCN(aq) + CsOH(aq) <------> CsCN(aq) + H2O(l)

Number of moles of HCN = (0.350mol/1000ml) ×50ml = 0.0175mol

Number of moles of CsOH required = 0.0175mol

Volume CsOH solution containing 0.0175mol of CsOH = (1000ml/0.250mol) × 0.0175mol = 70ml

Equivalence point = 70ml

At equivalence point all the HCN converted to CN^-

Total volume at equivalence point = 50ml + 70ml = 120ml

dilution factor of CN^- = 120ml /50ml = 2.4ml

[CN^-] at equivalence point = 0.350M /2.4 = 0.1458M

CN^- partly hydrolysed by water

CN^-(aq) + H2O(l) ------> HCN(aq) + OH^-(aq)

Kp = [HCN][OH^-]/ [CN^-]

Kp = Kw/Ka = 1^-].00×10^-14/4.0×10^-10 = 2.5×10^-5

at equillibrium

[CN^- ] = 0.1458 -x

[HCN] = x

[OH^- ]= x

x²/(0.1458 -x) = 2.5× 10^-5

solving for x

x = 0.001897

[OH^-] = 0.001897M

pOH = - log[OH^-]

pOH = -log(0.001897)

pOH = 2.72

pH = 14 -pOH

pH = 14 - 2.72

pH = 11.28