Question

a. Enough of a monoprotic acid is dissolved in water to produce a 0.0163 M solution. The pH of the resulting solution is 2.51. Calculate the Ka for the acid.

Ka= ________

b. The Ka of a monoprotic weak acid is 6.61 × 10-3. What is the percent ionization of a 0.108 M solution of this acid?

Percent ionization= ___________

Answer #1

a) moprotic acid dissociate as

HA + H_{2}O
H_{3}O^{+} + A^{-}

K_{a} = [H_{3}O^{+}] [
A^{-}] / [HA]

[H_{3}O^{+}] = 10^{-pH} = 0.00309 M

[H_{3}O^{+}] = [A^{-}] = 0.00309 M

substitute thsse value

K_{a} = (0.00309) (0.00309) /
0.0163

K_{a} = 0.000586 = 5.86
10^{-4}

b)

Weak acid dissociates as AH ⇌ A- + H+

K_{a} = [A^{-}][H^{+}] / [AH]

but [A^{-}] = [H^{+}] = x

K_{a} = [x][x] / [AH]

Substitute the value in equation

6.6110^{-3} =
[x]^{2}/ 0.108

[x]^{2} = 6.6110^{-3}0.108 =
0.00071388

[x] = 0.0267185

Concentration of H^{+} = 0.0267 M

% dissociation = [H^{+}] 100 / [HA]

% dissociation = 0.0267 100 / 0.108 = 24.72 %

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Hi can someone please help me do this question with explnation
and step by step.
Thank you

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