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_{a}V_{a} = M_{b}V_{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} = x^{2} / 0.25-x

6.166x10^{-10} * 0.25 = x^{2}

x = [H^{+}] = 1.2416x10^{-5} M

pH = -log(1.2416x10^{-5})

pH = 4.91

Hope this helps

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