Problem Set #9, question #5...

5. You would like to perform a titration using cyanic acid { HCNO(aq), K_a = 2x10^-4 } and potassium hydroxide. Which solution should be in the buret? You do not have a pH meter, but you have found a number of indicators. Which of the following would be most appropriate for this titration and why? Indicator: Endpoint range - 2,4-dinitrophenol: 2.8-4.0 ; phenol red: 6.7-8.1 ; thymolphthalein: 9.6-10.3

Although this question has some numbers in it, you don't actually need to do any calculations to answer it. Based upon the Ka value, HCNO(aq) is a weak acid. When performing a titration, you always want to put something strong and monoprotic in the buret. HCNO(aq) is monoprotic, but it is weak. Potassium hydroxide is monoprotic and strong, so it would be the better choice to put in the buret. Now, what about the indicator? When titrating a weak monoprotic acid with a strong monoprotic base, the equivalence point should be somewhere on the basic side, so we should choose an indicator that changes color at basic pH. The range for phenol red includes some slightly basic pH's so it should work, but it's pretty close to neutral, so thymolphthalein is probably the best bet for this titration.

We can also look at this by doing a calculation. When we get to the equivalence point, we will (in essence) have a solution of potassium cyanate in water. If we assume that the concentration of cyanate is 1M, we can calcuate an expected pH of the solution using a K_b-type of equation. {I just pulled 1M out of my hat, you can do this calculation with any concentration to check...}

CNO^-(aq) + H2O(l) <=> HCNO(aq) + OH^-(aq)

[CNO^-]_initial = 1 M

[HCNO]_initial = 0 M

[OH^-]_initial = 10^-7 M

[CNO^-]_change = -x M

[HCNO]_change = +x M

[OH^-]_change = +x M

[CNO^-]_eq = (1-x) M

[HCNO]_eq = x M

[OH^-]_eq = (10^-7+x)M

Let's assume that x is significantly smaller than 1 and significantly larger than 10^-7, then our K_b expression is:

K_b = {[HCNO]_eq [OH^-]_eq} / [CNO^-]_eq = { (x)(x) } / 1 = x² = 5x10^-11

NOTE: K_b for CNO^- is calculated from the given K_a of HCNO

x = 7.07x10^-6

Hmm, this is a pretty small number, our assumption that x is significantly larger than 10^-7 might be a little questionable. 10^-7 is only about 1.4% of 7.07x10^-6, so we should be OK...

x = 7.07x10^-6 = [OH^-]eq

pOH = -log [OH^-] = -log(7.07x10^-6) = 5.15

At 25 degC,

pH = 14 - pOH = 8.849

So it looks like the pH at the equivalence point should be basic enough that thymolphthalein will be a better indicator. For practice, redo this calculation assuming that the concentration of potassium cyanate is 0.5M at the equivalence point...