Question

Suppose you titrate 20.00 mL of a 0.250 M HCN solution using 0.100 M NaOH solution. (ka= 6.166 * 10^-10) (A) At what volume of base added will you have reached the equivalence point? (B) At what volume of base added will you have reached your half-way point? What is the pH of the solution at this point? (C) What is the pH of the solution prior to addition of any base? (D) Using excel or a similar program, determine the pH of the solution after the addition of 5.00 mL increments of NaOH solution, from 0-75.00 mL. Write the pH's below. Make a properly formatted titration curve and attach it.

Answer 1

I will answer till C, try to do D) yourself in excel. You already have the concentrations.

a) to reach the equivalence point:
M_aV_a = M_bV_b
V_b = 0.250 * 20 / 0.1 = 50 mL of base

b) the halfway point is reached when only the half of the volume required to reach the equivalence point is reached. In this case it would be: 50/2 = 25 mL

At this point, only half of the base is added, so half of the acid is consumed. Therefore the ratio of acid and base, would be 1, and the pH = pKa

pH = -log(6.166x10^-10) = 9.21

c) The pH of the solution prior to addition of base will be:
r: HCN --------> CN^- + H⁺
i: 0.250 0 0
e: 0.25-x x x

6.166x10^-10 = x² / 0.25-x
6.166x10^-10 * 0.25 = x²
x = [H⁺] = 1.2416x10^-5 M
pH = -log(1.2416x10^-5)
pH = 4.91

Hope this helps