Question

In: Chemistry

Enough of a monoprotic acid is dissolved in water to produce a 0.0103 M solution. The...

Enough of a monoprotic acid is dissolved in water to produce a 0.0103 M solution. The pH of the resulting solution is 2.53. Calculate the Ka for the acid.

I got 8.45x10^-4 but it says its wrong.

Expert Solution

Answer: Let the mono protic acid is HA.

Given pH of the solution = 2.53

HA when dissolved in water forms solution.

HA +H₂O ? H₃O⁺ + A^-

Ka = [H₃O⁺][A^-]/[HA]

HA + H₂O ? H₃O⁺ + A^-

Intial concentration: 0.0103 0 0

At equilibrium concentration: (0.0103-x) x x

x = [H₃O⁺]

pH = -log[H₃O⁺]

log[H₃O⁺]= -pH

= -2.53

[H₃O⁺] = 10^-2.53

=2.95*10^-3

Thus x = 2.95*10^-3

So:

Ka = [H₃O⁺][A^-]/[HA]

= [x][x]/[0.0103-x]

= [2.95*10^-3][2.95*10^-3]/[0.0103-2.95*10^-3]

= (8.7)*10^-6/7.35*10^-3

= 1.18*10^-3

Ka of the acid = 1.18*10^-3

queen_honey_blossom answered 2 hours ago

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