Answer to Question #245269 in General Chemistry for Sabra

Question

Calculate the pH of a solution prepared by mixing 2 ml of a strong acid solution of pH 3 and 3 ml of a strong base of pH 10

Answer 1

pH of Strong Acid = 3

-log[H⁺] = 3

[H⁺] = 10^,-3

pH of strong Base = 10

$\therefore$ pOH of strong base = 14 - 10 = 4

log[OH^-] = 4

[OH^-] = 10^-4

Moles of H⁺ in 2 ml solution = 2 × 10^-3 × 10^-3

Moles of OH^- in 3 ml solution = 3 × 10^-4 × 10^-3

Now, Reaction

$H^++OH^- \quad^{\rightarrow }_\leftarrow \quad H_2O_{(l)}$

Moles of H⁺ remaining in (3+2)ml solution = (2*l×10^-6 - 3×10^-7) = 1.7 x 10^-6

Molarity of H+ in solution = $\dfrac{1.7×10^{-6}}5×1000 = 3.40× 10^{-4}$

[H+] = 3·40 x 10^-4

pH = -log[H⁺]

pH = -log [3-40х10^-4]

pH = 3.46 $\approx$ 3.5