Calculate the pH

Question

An analytical chemist is titrating 241.4 mL of a 0.2300M solution of cyanic acid (HCNO) with a 1.200M solution of NaOH. The pK_a of cyanic acid is 3.46.

Calculate the pH of the acid solution after the chemist has added 22.78 mL of the NaOH solution to it.

you may assume the final volume equals the initial volume of the solution plus the volume of NaH solution added.

Round your answer to 2 decimal places.

Answer 1

When adding NaOH to a weak acid such as HCNO, a buffer will be formed since you are creating the salt of that weak acid. Up to the point of equivalence, you will then have the presence of BOTH the weak acid, and the salt of that acid (conjugate base). This is the definition of a buffer. We can use the Henderson Hasselbalch equation for find the pH of such a buffer solution.

HCNO + OH^- ==> H2O + CNO^- (omitted Na since it is merely a spectator ion)

moles of HCNO initially present = 241.1 ml x 1 L / 1000 ml x 0.2300 mol / L = 0.05545 mols

moles OH- added = 22.78 ml x 1 L / 1000 ml x 1.200 mol / L = 0.02734 moles

HCNO + OH^- ==> H2O + CNO^-

0.05545...0.02734......0.........0..............Initial

-0.02734...-0.02734...........+0.02734....Change

0.02811.........0...................0.02734......Equilibrium

Final volume of solution = 241.1 ml + 22.78 ml = 263.9 mls

Final [HCNO] = 0.02811 mol / 0.2639 L = 0.1065 M

Final [CNO-] = 0.02734 mol / 0.2639 L = 0.1036 M

Henderson Hasselbalch: pH = pKa + log [salt] / [acid]

pH = 3.46 + log (0.1036 / 0.1065)

pH = 3.46 - 0.012

pH = 3.45

(be sure to check all of the math)