Question

A solution is prepared by combining 38.614 mL of 2.10 M aqueous weak acid and 38.614 mL of 1.05 M aqueous NaOH. What is the pH of the resulting solution? The K_a of the weak acid is 8.844 x 10^-8. Enter your answer with at least two decimal places.

Please show your work for how you got your answer.

Answer 1

Answer:

Volume of acid = 38.614 mL = 38.614 × 10^-3 L

Concentration of acid = 2.10M

K_a = 8.844 × 10^-8

[H⁺]= (K_a×C)^1/2

[H⁺]= (8.844 × 10^-8×2.10)^1/2

[H⁺]= 4.309 × 10^-4 M

No. of equivalents of H⁺ = [H⁺]× V

No. of equivalents of H⁺ = 4.309 × 10^-4 × 38.614 × 10^-3

No. of equivalents of H⁺ =1.664 × 10^-5

Volume of NaOH = 38.614 mL = 38.614 × 10^-3 L

Concentration of acid = 1.05 M

No. of equivalents of OH^- = C× V

No. of equivalents of OH^- = 1.05 × 38.614 × 10^-3

No. of equivalents of OH^- =4.045 × 10^-2

Out of which,1.664 × 10^-5 OH^- ions will be neutralised by H+ ions.

OH^- ions remaining in the solution = 4.045 × 10^-2 - 1.664 × 10^-5

OH^- ions remaining in the solution = 0.04053121

_POH = -log[OH^-]

_POH = -log[0.04053121]

_POH = 1.3922

_PH = 14-_POH

_PH = 14-1.3922

_PH = 12.60