JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
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 \(x \times 10^{-4} \mathrm{M}\). The value of x is ________ (nearest integer)
- A 20
- B 25
- C 30
- D 35
Answer & Solution
Correct Answer
(B) 25
Step-by-step Solution
Detailed explanation
\begin{aligned} & \quad \mathrm{HX}_{(\mathrm{aq)}} \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{X}_{(\mathrm{aq})}^{-} \quad \mathrm{K}_{\mathrm{a}}=4 \times 10^{-10} \\ & 0.01(1-\alpha) \quad 0.01 \alpha \quad 0.01 \alpha \quad \text { Not justified } \\ &…
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