26 ml of 0037 M hydroiodic acid is added to 677 mt of 0 074

26 ml. of 0.037 M hydroiodic acid is added to 677 mt of 0 074 M hydrochioni acid Callate the pH of the uting solution of srng od 2 Answer 1.13 1.84 1.22 143 1.28

Solution

We are given a mixture of two strong acids (Hydroiodic acid and Hydrochloric acid). Thus,

M₁V₁ + M₂V₂ = M(V₁ + V₂)

We are given,

M₁ (Hydroiodic acid) = 0.037 M

V₁ = 426 mL

M₂ (Hydrochloric acid) = 0.074 M

V₂ = 677 mL

On substituting given values in above equation,

M₁V₁ + M₂V₂ = M(V₁ + V₂)

(0.037 M) * (426 mL) + (0.074 M) * (677 mL) = M * (426 mL + 677 mL)

15.762 + 50.098 = M * (1103 mL)

M = 5.97 * 10^-2

pH = - log [M]

pH = - log (5.97 * 10^-2 )

pH = - log 5.97 + 2 log10

pH = - 0.776 + 2

pH = 1.224