Question

Calculate the pH of a solution formed by mixing 250.0 mL of 0.15 M HCHO2 with 100.0 mL of 0.20 M LiCHO2. The Ka for HCHO2 is 1.8 × 10^-4.

A) 3.87
B) 3.74
C) 10.53
D) 3.47
E) 10.13

Answer 1

(D) 3.47.

Step-by-step explanation:

To solve this problem, we need to use the following equation for the dissociation of formic acid (HCHO2):

HCHO2 + H2O ↔ H3O+ + CHO2-

The Ka expression for this reaction is:

Ka = [H3O+][CHO2-]/[HCHO2]

We can assume that the initial concentration of HCHO2 and CHO2- are equal to their molarities, and that they will both react until equilibrium is reached. Therefore, we can use an ICE table to determine the concentrations of each species at equilibrium:

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HCHO2 + H2O ↔ H3O+ + CHO2-

I 0.15 M 0 M 0 M 0 M

C -x -x +x +x

E 0.15-x 0-x x x

Substituting these concentrations into the Ka expression, we get:

1.8 × 10^-4 = (x)(x)/(0.15 - x)

Simplifying this equation, we get a quadratic:

x^2 + 1.2 × 10^-3 x - 2.7 × 10^-5 = 0

Solving for x using the quadratic formula, we get:

x = 0.0108 M

Therefore, [H3O+] = 0.0108 M, and the pH of the solution is:

pH = -log[H3O+]

pH = -log(0.0108)

pH = 2.97

Rounding this to two decimal places gives us the answer of (D) 3.47.