The enthalpy of Neutralisation of strong acids HA and weak acid HB by NaOH are -13.7 and -12.7 $K . C a l e q^{- 1}$ .When one equivalent of NaOH is added of HA and HB, the enthalpy change was $- 13.5 K . c a l$ . In what ratio is the base distributed between HA and HB ?

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2:1

1:2

4:1

1:4

answer is C.

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Detailed Solution

Let the base be distributed between HA and HB in the following ratio a:b

$\begin{array}{l} a + b = 1 - - - - 1 \\ But given that HA + NaOH \to H_{2} O + NaA; Δ H = - 13.7 K .cal \\ HB + NaOH \to H_{2} O + NaB; Δ H = - 12.7 K . cal \\ a (- 13.7) + b (- 12.7) = - 13.5 \\ 13.7 a + 12.7 b = 13.5 - - - 2 \\ From 1 & 2 b = 0.2; a = 0.8 \\ \frac{a}{b} = \frac{0.8}{0.2} = \frac{4}{1} (or) 4 : 1 \end{array}$