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Question: In a mixture of acetic acid and sodium acetate, the ratio of concentrations of the salt to the acid ...

In a mixture of acetic acid and sodium acetate, the ratio of concentrations of the salt to the acid is increased ten times. Then the pH of the solution:
A. Increases by one
B. Decreases by one
C. Decreases ten fold
D. Increases tenfold

Explanation

Solution

When the ratio of the concentrations of the salt to the acid is increased ten times, the pH increases by one. The pH of any salt solution is equal to the sum of the pKa and the log of the ratio of the concentrations of the salt to the acid in the solution

Complete answer:
First, let us understand the reaction occurring in the solution.
CH3COOH+NaOHCH3COONa+H2OC{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O
As we can clearly see in such a neutralization reaction where a weak acid like acetic acid is added to a strong base like sodium hydroxide and gets neutralized by the formation of a salt and the water.
Now, applying the formula of the pH involved during such reactions, we have:
pH=pKa+logc(BaseAcid)pH = p{K_a} + \log_c \left( {\dfrac{{Base}}{{Acid}}} \right)
In the question, as it is said that the ratio of the concentrations of the salt to the acid in the solution is 10. Now applying this value we have:
[Base][Acid]=10 pH=pKa+log(10) pH=pKa+1 \begin{gathered} \dfrac{{[Base]}}{{[Acid]}} = 10 \\\ \therefore pH = p{K_a} + \log (10) \\\ pH = p{K_a} + 1 \\\ \end{gathered}

So, the correct answer is Option A.

Note:
In the given question one must keep in mind that the ratio of the conjugate base (salt) and the acid has been provided. If it would have been vice versa, then the positive sign between pKa and the logarithmic part in the mathematical formula would have been reversed and would have become negative.