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Question: A solution of weak acid, \({\text{HA}}\), has a concentration of \(0.100{\text{ M}}\). What is the c...

A solution of weak acid, HA{\text{HA}}, has a concentration of 0.100 M0.100{\text{ M}}. What is the concentration of hydronium ion and the pH of this solution if the Ka{{\text{K}}_{\text{a}}} value for this acid is 1.0×10−51.0 \times {10^{ - 5}}?
(A) 1.0×10−3 and pH=111.0 \times {10^{ - 3}}{\text{ and pH}} = 11
(B) 1.0×10−6 and pH=61.0 \times {10^{ - 6}}{\text{ and pH}} = 6
(C) 1.0×10−4 and pH=81.0 \times {10^{ - 4}}{\text{ and pH}} = 8
(D) 3.0×10−4 and pH=43.0 \times {10^{ - 4}}{\text{ and pH}} = 4
(E) 1.0×10−3 and pH=31.0 \times {10^{ - 3}}{\text{ and pH}} = 3

Explanation

Solution

To solve this first write the dissociation reaction of weak acid. Then calculate the concentration of the hydronium ion using the equation of acid dissociation constant. Then after that we can proceed to calculate the value of the pH of the solution..

Complete step by step solution: We are given a solution of weak acid, HA{\text{HA}}, which dissociates into hydronium ion and an anion. The dissociation reaction of weak acid HA{\text{HA}} is as follows:
HA(aq)⇌H3O+(aq)+A−(aq){\text{HA}}\left( {{\text{aq}}} \right) \rightleftharpoons {{\text{H}}_3}{{\text{O}}^ + }\left( {{\text{aq}}} \right) + {{\text{A}}^ - }\left( {{\text{aq}}} \right)
The expression for acid dissociation constant for the given weak acid is as follows:
Ka=[H3O+][A−][HA]{{\text{K}}_{\text{a}}} = \dfrac{{[{{\text{H}}_3}{{\text{O}}^ + }][{{\text{A}}^ - }]}}{{[{\text{HA}}]}} …… (1)
Where, Ka{{\text{K}}_{\text{a}}} is the acid dissociation constant.
The initial concentration of HA{\text{HA}} is given as 0.100 M0.100{\text{ M}} and the initial concentrations of H3O+{{\text{H}}_3}{{\text{O}}^ + } and A−{{\text{A}}^ - } are zero.
Let the final concentration of H3O+{{\text{H}}_3}{{\text{O}}^ + } be x and A−{{\text{A}}^ - } be x. Thus, the final concentration of HA{\text{HA}} is 0.100−x M0.100 - x{\text{ M}}
Thus, equation (1) becomes,
1.0×10−5=x⋅x0.100−x1.0 \times {10^{ - 5}} = \dfrac{{x \cdot x}}{{0.100 - x}}
The value of x is very small. Thus, x is negligible. Thus, 0.100−x=0.1000.100 - x = 0.100. Thus,
1.0×10−5=x⋅x0.1001.0 \times {10^{ - 5}} = \dfrac{{x \cdot x}}{{0.100}}
⇒x2=1.0×10−5×0.100\Rightarrow {x^2} = 1.0 \times {10^{ - 5}} \times 0.100
⇒x2=1.0×10−6\Rightarrow {x^2} = 1.0 \times {10^{ - 6}}
⇒x2=1.0×10−3\Rightarrow {x^2} = 1.0 \times {10^{ - 3}}
Now, we know that x=[H3O+]x = [{{\text{H}}_3}{{\text{O}}^ + }]. Thus,
[H3O+]=1.0×10−3[{{\text{H}}_3}{{\text{O}}^ + }] = 1.0 \times {10^{ - 3}}
Thus, the concentration of hydronium ion in the solution of weak acid is 1.0×10−31.0 \times {10^{ - 3}}.
We know that the negative logarithm of the hydrogen ion or hydronium ion concentration is known as pH. Thus,
pH=−log⁡[H3O+]{\text{pH}} = - \log [{{\text{H}}_3}{{\text{O}}^ + }]
Substitute 1.0×10−31.0 \times {10^{ - 3}} for the concentration of hydronium ion and solve for the pH. Thus,
pH=−log⁡[1.0×10−3]{\text{pH}} = - \log [1.0 \times {10^{ - 3}}]
⇒pH=3\Rightarrow {\text{pH}} = 3
Thus, the pH of a solution of weak acid is 3.
Thus, a solution of weak acid, HA{\text{HA}} having a concentration of 0.100 M0.100{\text{ M}} has the concentration of hydronium ion 1.0×10−31.0 \times {10^{ - 3}} and the pH of this solution is 3.

Thus, the correct option is (E) 1.0×10−3 and pH=31.0 \times {10^{ - 3}}{\text{ and pH}} = 3.

Note: We are given a weak acid. The another formula to calculate the pH of a weak acid directly from the given value of dissociation constant of acid is as follows:
pH=12pKa−log⁡[acid]{\text{pH}} = \dfrac{1}{2}{\text{p}}{{\text{K}}_{\text{a}}} - \log [{\text{acid}}]