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Question: The pH of \(0.15M\) solution of \(HOCl\) \(({K_a} = 9.6 \times {10^{ - 6}})\) is: A. \(4.42\...

The pH of 0.15M0.15M solution of HOClHOCl (Ka=9.6×10−6)({K_a} = 9.6 \times {10^{ - 6}}) is:
A. 4.424.42
B. 2.922.92
C. 3.423.42
D. None of these

Explanation

Solution

We know that HOClHOCl is a weak acid so HOClHOCl will dissociate to give small amount of H3O+{H_3}{O^ + } and A−{A^ - } . The equilibrium constant for an acid is called the acid ionization constant, KaKa . First we will determine the concentration of H+{H^ + }ions present in the solution using the formula KaC\sqrt {KaC} where KaKa is the equilibrium constant and CC is the concentration of the acid. After finding the H+{H^ + } ions we will find the pH by using the formula −log⁡[H+] - \log [{H^ + }] .

Complete step by step answer:
First, we will write the equilibrium equation:
HOCl(aq)+H2O(l)⇌H3O+(aq)+Cl−(aq)HOCl(aq) + {H_2}O(l) \rightleftharpoons {H_3}{O^ + }(aq) + C{l^ - }(aq)
We know that for acidic solutions
H+=KaC{H^ + } = \sqrt {{K_a}C}
Where KaKa =9.6×10−6 = 9.6 \times {10^{ - 6}} and C=0.15MC = 0.15M
On putting the values of KaKa and CC in the above formula we get: -
H+=KaC{H^ + } = \sqrt {{K_a}C}
=9.6×10−6×0.15 =1.44×10−6 =0.0012M  = \sqrt {9.6 \times {{10}^{ - 6}} \times 0.15} \\\ = \sqrt {1.44 \times {{10}^{ - 6}}} \\\ = 0.0012M \\\
We got the value of H+{H^ + }so now we can simply determine the pH of the solution by using the formula: -
pH=−log⁡[H+]pH = - \log [{H^ + }]
By putting the value of H+{H^ + } in this formula we get: -
pH=−log⁡[H+]pH = - \log [{H^ + }]
=−log⁡[0.0012] =2.920818  = - \log [0.0012] \\\ = 2.920818 \\\
i.e.,option (B) is correct.

Additional information: By determining the equilibrium constants in aqueous solutions, we can measure the relative strength of acid. The larger the KaKa of an acid, the larger will be the concentration of H3O+{H_3}{O^ + } and A−{A^ - } . Thus, the ionization constant increases as the strength of the acid increases. The concentration of the H3O+ ion in an aqueous solution gradually decreases and the pH of the solution increases as the solution becomes more dilute.

Note: : The two assumptions that are made in weak-acid equilibrium are as follows:
1-The dissociation of the acid is so small that the change in the concentration of the acid can be ignored.
2-The dissociation of the acid is so large that the H3O+ ion concentration can be ignored.