What is the pH after.05 M 200ml of NaOH is added to 200mL of .05 M HCN (Ka=4.9x10^-10)

Question

What is the pH after.05 M 200ml of NaOH is added to 200mL of .05 M HCN (Ka=4.9x10^-10)

Answer 1

200ml of .05M NaOH will neutralize 200 ml of .05 M HCN leaving 400 ml of .025 M CN^-

Now you can adjust for equilibrium of .025M CN^-

CN^- +H₂O ⇔ HCN + OH^- K_B = K_W/K_A = 10^-14/(4.9 x 10^-10 = 2.0408 x 10^-6

Let x be the amount of CN^- that goes back to HCN

K_B = 2.0408 x 10^-6 = x(10^-7+x)/(.025-x) but x<< .025 and x >> 10^-7 (assume)

2.0408 x 10^-6 = x²/.025 and x = 2.259 x 10^-4 ([OH^-] = 2.260 x 10^-4 M {well within data sig figs}

pH = 14 - log[OH^-]

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