10 ml of weak acid (HA) is 20% dissociated in water. The acid solution required 10 ml of 2×10−3M NaOH solution to reach equivalence point. Which of the following is/are true? (log 4 = 0.6)

m. eq of NaOH=2×10−3×10=0.02=m.eq HA NHA=0.0210=2×10−3 [H+]=2×10−3×0.2=4×10−4 pH=3.4 Ka=α21−α=2×10−3(0.2)2(1−0.2) Ka=10−4 HA+NaOH→NaA(basic salt) pH=7+12(pKa+logC) [salt]=0.0220=10−3=7+12(4+log10−3)=7.5 Kh=KwKa=10−1410−4=10−10

10 ml of weak acid (HA) is 20% dissociated in water. The acid solution required 10 ml of 2×10−3M NaOH solution to reach equivalence point. Which of the following is/are true? (log 4 = 0.6)

m. eq of NaOH=2×10−3×10=0.02=m.eq HA NHA=0.0210=2×10−3 [H+]=2×10−3×0.2=4×10−4 pH=3.4 Ka=α21−α=2×10−3(0.2)2(1−0.2) Ka=10−4 HA+NaOH→NaA(basic salt) pH=7+12(pKa+logC) [salt]=0.0220=10−3=7+12(4+log10−3)=7.5 Kh=KwKa=10−1410−4=10−10

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10 ml of weak acid (HA) is 20% dissociated in water. The acid solution required 10 ml of
$2 \times 10^{- 3} M$ NaOH solution to reach equivalence point. Which of the following is/are true? (log 4 = 0.6)

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pH of the weak acid is 3.4

Ka of the weak acid is $10^{- 4}$

pH of the solution at the equivalence point is 7.5

$K_{h}$ of the salt NaA is $10^{- 10}$

answer is A, B, C, D.

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Detailed Solution

m. eq of $NaOH = 2 \times 10^{- 3} \times 10 = 0.02 = m.eq HA$
$N_{H A} = \frac{0.02}{10} = 2 \times 10^{- 3}$
$[H^{+}] = 2 \times 10^{- 3} \times 0.2 = 4 \times 10^{- 4}$
$p H = 3.4$
$K a = \frac{α^{2}}{1 - α} = \frac{2 \times 10^{- 3} {(0.2)}^{2}}{(1 - 0.2)}$
$K a = 10^{- 4}$
$H A + N a O H \to N a A (basic salt)$
$p H = 7 + \frac{1}{2} (p^{K a} + \log C) [salt] = \frac{0.02}{20} = 10^{- 3} = 7 + \frac{1}{2} (4 + \log 10^{- 3}) = 7.5$
$K h = \frac{K w}{K a} = \frac{10^{- 14}}{10^{- 4}} = 10^{- 10}$