0.5 M HA and 0.4 M HB are present in two different solutions. The volume of solution containing HA is 10 L whereas volume of solution containing HB is 30 L . The dissociation constants of both the weak acids HA and HB are $3 \times 10^{- 4}$ and $5 \times 10^{- 4}$ respectively.
The pH of solution at equilibrium after both HA and HB solutions are mixed together is :

8.46

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1.86

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4.98

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6.05

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Solution

The correct option is B 1.86
$C_{a} = 0.5 M C_{b} = 0.4 M V_{a} = 10 L V_{b} = 30 L$

Concentration of HA after mixing= $\frac{C_{a} V_{a}}{V_{a} + V_{b}} = \frac{0.5 \times 10}{10 + 30} = 0.125 M = C_{1}$

Concentration of HB after mixing= $\frac{C_{b} V_{b}}{V_{a} + V_{b}} = \frac{0.4 \times 10}{10 + 30} = 0.1 M = C_{2}$

$H A (a q .) ⇌ H^{+} (a q .) + A^{-} (a q .) I n i t i a l : C_{1} 00 E q u i l i b r i u m : C_{1} (1 - α_{1}) C_{1} α_{1} + C_{2} α_{2} C_{1} α_{1}$

$H B (a q .) ⇌ H^{+} (a q .) + B^{-} (a q .) I n i t i a l : C_{2} 00 E q u i l i b r i u m : C_{2} (1 - α_{2}) C_{1} α_{1} + C_{2} α_{2} C_{2} α_{2}$

$p H = - \frac{1}{2} l o g [(K_{a 1} C_{1}) + (K_{a 2} C_{2})] p H = - \frac{1}{2} l o g [(3 \times 10^{- 4} \times 0.125) + (5 \times 10^{- 4} \times 0.1)] p H = - \frac{1}{2} l o g (8.75 \times 10^{- 5}) p H = \frac{1}{2} (5 - l o g 8.75) = \frac{1}{2} (5 - 0.94) p H = 2.03$

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