Question

The pH of a 0.01 M weak acid $\mathrm{HX}\left(\mathrm{K}_{\mathrm{a}}=4 \times 10^{-10}\right)$ is found to be 5 . Now the acid solution is diluted with excess of water so that the pH of the solution changes to 6 . The new concentration of the diluted weak acid is given as $\mathrm{x} \times 10^{-4} \mathrm{M}$. The value of x is _______ (nearest integer).

Hint:

Check the consistency of the given data and ensure the calculations are justified.

Answer 1

Correct Answer: 25

This problem calculates the concentration of a diluted weak acid solution. Initially, the solution has a concentration of 0.01 M and a pH of 5.

A pH of 5 signifies an H₃O⁺ concentration of 10^-5 M. For a weak acid HX, the dissociation equilibrium is described by the expression: $$\mathrm{K}_a = \frac{[\text{H}_3\text{O}^+][\text{X}^-]}{[\text{HX}]}$$

Given $\mathrm{K}_a=4 \times 10^{-10}$ and [H₃O⁺] = 10^-5 M, the initial concentration of $\text{HX}$ is found to be 0.01 M.

At equilibrium, [H₃O⁺]=[X^-]=10^-5 M, and $\text{HX}$ remains approximately 0.01 M. This is consistent with the equilibrium constant, as $\mathrm{K}_a(=\frac{(10^{-5})(10^{-5})}{0.01})=10^{-10}$.

The solution is then diluted to a pH of 6, resulting in an H₃O⁺ concentration of 10^-6 M.

Using the $\mathrm{K}_a$ expression with the new H₃O⁺ concentration: $$\mathrm{K}_a = \frac{(10^{-6})^2}{\text{new [HX]}} = 4 \times 10^{-10}$$

Solving for the new concentration of $\text{HX}$: $$\text{new [HX]} = \frac{(10^{-6})^2}{4 \times 10^{-10}} = 2.5 \times 10^{-3} \mathrm{M}$$

The concentration is to be expressed in the format $\mathrm{x} \times 10^{-4} \mathrm{M}$. Converting the calculated concentration: $2.5 \times 10^{-3} \mathrm{M}$ is equivalent to $25 \times 10^{-4} \mathrm{M}$.

Therefore, the value of x is 25. This value falls within the specified range (25, 25), confirming it satisfies the condition.

The nearest integer for x is 25.