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JEE Advanced · Chemistry · 8. Ionic Equilibrium

At \(25^{\circ} \mathrm{C}\), the concentration of \(\mathrm{H}^{+}\)ions in \(1.00 \times 10^{-3} \mathrm{M}\) aqueous solution of a weak monobasic acid having acid dissociation constant \(\left(K_a\right)=4.00 \times 10^{-11}\) is \(\boldsymbol{X} \times 10^{-7} \mathrm{M}\). The value of \(\boldsymbol{X}\) is _____ .
Use: Ionic product of water \(\left(K_w\right)=1.00 \times 10^{-14}\) at \(25^{\circ} \mathrm{C}\)

  1. A 2.24
  2. B 2.54
  3. C 2.64
  4. D 2.74
Verified Solution

Answer & Solution

Correct Answer

(A) 2.24

Step-by-step Solution

Detailed explanation

Because concentration of \(\mathrm{H}+\) from weak acid is less we need to consider self ionization of \(\mathrm{H}_2 \mathrm{O}\) also.
\(\begin{array}{rl}\mathrm{HX}(\mathrm{aq}) \rightleftharpoons & \mathrm{H}^{+}+\mathrm{X}^{-}(\mathrm{aq}) \\ 10^{-3}-\mathrm{x} \quad & \mathrm{x}+\mathrm{y} \quad \mathrm{x} \\ \mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons & \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \\ & \mathrm{x}+\mathrm{y} \quad \quad \mathrm{y}\end{array}\)
Approximation : \(\left(10^{-3}-\mathrm{x}\right) \simeq 10^{-3}\)
\(\begin{aligned}
& \Rightarrow \quad \frac{x(x+y)}{10^{-3}}=K_a=4 \times 10^{-11}....(1) \\
& \Rightarrow \quad y(x+y)=K w=10^{-14}.....(2)
\end{aligned}\)
\(\begin{array}{ll}
\text { Add }(1)+(2) \\
\end{array}\)
\(\begin{array}{ll}\Rightarrow & (\mathrm{x}+\mathrm{y})^2=5 \times 10^{-14} \\ \Rightarrow & \mathrm{x}+\mathrm{y}=\left[\mathrm{H}^{+}\right]=\sqrt{5} \times 10^{-7} \\ \Rightarrow & \mathrm{x}=\sqrt{5}=2.236\end{array}\)
Answer 2.23 or 2.24.
From JEE Advanced
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