The pH of a 0.01 M weak acid HX (Ka = 4 × 10^(-10)) is found to be 5. Now the acid solution is diluted with excess of water ​

Question

The pH of a 0.01 M weak acid HX \((K_a = 4 \times 10^{-10} )\) 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 \(x \times 10^{-4} \ M\) The value of x is _________ (nearest integer)

Answer 1

Answer is: 25

After dilution

\(pH = 6 \quad [H^+] = 10^{-6}\)

\([H^+] = 10^{-6 } = \sqrt{K_a C}\)

\(10^{-6} = \sqrt{4 \times 10^{-10} \times C}\)

\(\frac{10^{-12}}{4 \times 10^{-10}} = C\)

\(0.25 \times 10^{-2} = C\)

\(20 \times 10^{-4} = C \quad \therefore \ x= 25\)