Question

Hz is a weak acid. an aqueous solution of hz is prepared by dissolving 0.020 mol of hz in sufficient water to yield 1.0 l of solution. the ph of the solution was 4.93 at 25.0 °c. the ka of hz is ________. hz is a weak acid. an aqueous solution of hz is prepared by dissolving 0.020 mol of hz in sufficient water to yield 1.0 l of solution. the ph of the solution was 4.93 at 25.0 °c. the ka of hz is ________. 1.4 ⋅ 10-10 1.2 ⋅ 10-5 2.8 ⋅ 10-12 9.9 ⋅ 10-2 6.9 ⋅ 10-9

Answer 1

The ka of hz is 6.9·10⁻⁹.
c(Hz) = n(Hz) ÷ V(Hz).
c(Hz) = 0.02 mol ÷ 1.0 L.
c(Hz) = 0.02 M.
pH = 4.93.
[H⁺]  = 10∧(-4.93) = 1.17·10⁻⁵ M.
Ka = (1.17·10⁻⁵ M)² / (0.02 M - 1.17·10⁻⁵ M).
Ka = 1.38·10⁻¹⁰ ÷ 0.0199 M.
Ka = 6.9·10⁻⁹.

Answer 2

From the information presented in the question, the Ka of the solution is 6.9 × 10^-9.

• First, we must set up the ICE table for the reaction;

Concentration of HZ = 0.020 mol/1L = 0.020 M

          HZ(aq) + H2O(l) ⇄ H3O^+(aq) +   Z^-(aq)

I          0.020                       0                   0  

C        -x                               +x                 +x

E      0.020 - x                     x                   x

• We obtain the Ka of the solution from;

Ka = [ H3O^+] [Z^-]/[HZ]

Ka = x^2/ 0.020 - x

• From the question, pH = 4.93

Hence,  [ H3O^+] = Antilog (-4.93) = 1.17 × 10^-5 M

Also;  [ H3O^+] = [Z^-] = x

Substituting values;

Ka = (1.17 × 10^-5)^2/ 0.020 - 1.17 × 10^-5

Ka = 6.9 × 10^-9

Hence, the Ka of the solution is 6.9 × 10^-9.

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