H2A is a weak diprotic acid. If the pH of 0.1 M H2A solution is 3 and concentration of A2- is 10-12 at 25°C. Select correct statement(s)

H2A⇌H++HA−∵y is very less so x+y≈x∴Ka1=x2c−xAt equilibrium C- X x+y x−y Ka2=y=10−12;pKa2=12HA−1⇌H++A2− at equilibrium x−y x+y ypKa2−pKa1=12−5=7pH=12pKa1−logC3=12pKa1+1pKa1=5

H2A is a weak diprotic acid. If the pH of 0.1 M H2A solution is 3 and concentration of A2- is 10-12 at 25°C. Select correct statement(s)

H2A⇌H++HA−∵y is very less so x+y≈x∴Ka1=x2c−xAt equilibrium C- X x+y x−y Ka2=y=10−12;pKa2=12HA−1⇌H++A2− at equilibrium x−y x+y ypKa2−pKa1=12−5=7pH=12pKa1−logC3=12pKa1+1pKa1=5

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H₂A is a weak diprotic acid. If the pH of 0.1 M H₂A solution is 3 and concentration of A^2- is 10^-12at 25°C. Select correct statement(s)

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$The value of K_{a_{1}} is nearly 10^{- 5}$

${[H^{+}]}_{Total} \approx [H^{+}] from first step of ionization of acid H_{2} A$

${pK}_{a_{2}} - {pK}_{1} = 9$

${Concentration of}^{-} OH in solution is 10^{- 3} M$

answer is A, C.

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Detailed Solution

$\begin{array}{l} H_{2} A ⇌ H^{+} + {HA}^{-} \\ ∵ y is very less so x + y \approx x ∴ K_{a_{1}} = \frac{x^{2}}{c - x} \end{array}$

At equilibrium

$C- X x + y x - y K_{a_{2}} = y = 10^{- 12}; p K_{a_{2}} = 12$

$\begin{array}{l} {HA}^{- 1} ⇌ H^{+} + A^{2 -} \\ at equilibrium x - y x + y y \\ {pK}_{a_{2}} - {pK}_{a_{1}} = 12 - 5 = 7 \\ pH = \frac{1}{2} [{pK}_{a_{1}} - \log C] \\ 3 = \frac{1}{2} [{pK}_{a_{1}} + 1] {pK}_{a_{1}} = 5 \end{array}$