Dear Student, Degree of dissociation (α) = λmcλm∞=10200=0 05and, for weak acidsKa=Cα21-α = 0 1×(0 05)21-0 05=0 000250 95=2 63×10-4Now, HA ↔ H+ + A-Ka=[H+][A-][HA]or Ka=[H+]2[HA]or, [H+]2= Ka[HA]Taking log both sides, log [H+]2= log Ka[HA]or, 2log[H+]=log Ka+log [HA]or, log[H+]=log Ka+log [HA]2or, -log [H+] =log [HA]-log Ka2or, pH=log (0 1)-log (2 63×10-4)2=-1-0 41+42=1 3Hence, correct answer is (1)

CORRECT OPTION?
CORRECT OPTION? Question 17 of 90 Bookmark I Flag Que Equivalent conductance 0.1 M H A (Weak acid ) solution is 10 Scm2 equivalentl and that at infinte dilution is 200 Scm2 equivalent-I Hence pH of HA Solution is 1.3 17 23 37

Dear Student,

Degree of dissociation (α) = \frac{{λ_{m}}^{c}}{{λ_{m}}^{\infty}} = \frac{10}{200} = 0.05 and, for weak acids K_{a} = \frac{{Cα}^{2}}{1 - α} = \frac{0.1 \times (0.05)^{2}}{1 - 0.05} = \frac{0.00025}{0.95} = 2.63 \times 10^{- 4} Now, HA \leftrightarrow H^{+} + A^{-} K_{a} = \frac{[H^{+}] [A^{-}]}{[HA]} or K_{a} = \frac{[H^{+}]^{2}}{[HA]} or, [H^{+}]^{2} = K_{a} [HA] Taking \log both sides, \log [H^{+}]^{2} = \log K_{a} [HA] or, 2 \log [H^{+}] = \log K_{a} + \log [HA] or, \log [H^{+}] = \frac{\log K_{a} + \log [HA]}{2} or, - \log [H^{+}] = \frac{\log [HA] - \log K_{a}}{2} or, pH = \frac{\log (0.1) - \log (2.63 \times 10^{- 4})}{2} = \frac{- 1 - 0.41 + 4}{2} = 1.3 Hence, correct answer is (1)

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​CORRECT OPTION?​​​CORRECT OPTION?​​ Question 17 of 90 Bookmark I Flag Que Equivalent conductance 0.1 M H A (Weak acid ) solution is 10 Scm2 equivalentl and that at infinte dilution is 200 Scm2 equivalent-I Hence pH of HA Solution is 1.3 17 23 37

CORRECT OPTION?
CORRECT OPTION? Question 17 of 90 Bookmark I Flag Que Equivalent conductance 0.1 M H A (Weak acid ) solution is 10 Scm2 equivalentl and that at infinte dilution is 200 Scm2 equivalent-I Hence pH of HA Solution is 1.3 17 23 37