$H_{0} is an acidic indicator (K_{\ln} = 10^{- 7}) which dissociates into H^{+} and {In}^{-} in$ aqueous solution. It is added to a solution of 30 ml of 0.15 M $H_{3} {PO}_{4}$
$(K_{1} = 1 \times 10^{- 3}, K_{2} = 1 \times 10^{- 7} \overset{- 1}{=} K_{3} = 1 \times 10^{- 13}) . If Hln and {In}^{-} posses colour P$ and Q respectively and color P predominates over color Q, when concentration of HIn is 120 times than that of $I n^{-}$ and color Q predominates over P, and when concentration of $I n^{-}$ is 127 times of HIn. If this solution treated with 30 ml KOH,
then correct statements among the following is/are :

Difficult

Solution

$\begin{array}{l} H_{3} {PO}_{4} & \overset{K_{1} = 10^{- 3}}{⇌} & H^{+} & + & H_{2} {PO}_{4}^{-} \\ t = 0 & C & 0 & 0 \\ t = t & C (1 - α) & C α & C α \end{array}$
$H^{+} = C α = \sqrt{K a . C} = \sqrt{1.5 \times 10^{- 3}} = 1.23 \times 10^{- 2}$
$K_{I n} = \frac{[H^{+}] [I n^{-}]}{[H_{I n}]}$
$10^{- 7} = 1.23 \times 10^{- 2} \times \frac{[I n^{-}]}{[H_{I n}]}$
$\frac{[H_{I n}]}{[I n^{-}]} = \frac{1.23 \times 10^{- 2}}{10^{- 7}} = 1.23 \times 10^{5}$
Now, when P colour predominates over Q colour, then
$p H = p K_{I n} + \log \frac{1}{120}$
$p H = 7 - \log 120 = 4.93$
When Q colour predominates over P colour, then
$p H = p K_{I n} + \log 127 = 9.1$
\ pH range of the indicator is 4.93 to 9.1
$At 1^{st} neutralization point, the acid H_{3} {PO}_{4} will be convert into {KH}_{2} {PO}_{4}, here pH$ of the solution
$= \frac{p K_{1} + p K_{2}}{2} = \frac{3 + 7}{2} = 5$
$At the 2^{nd} step of neutralization, the acid H_{3} {PO}_{4} will be convert into K_{2} {HPO}_{4},$
here the pH of the solution $= \frac{p K_{2} + p K_{3}}{2} = \frac{7 + 13}{2} = 10$
The equivalence point will be obtained when acid $H_{3} P O_{4}$ will change into $K_{2} H P O_{4}$
\n- factor of the acid = 2
\normality of $K O H = \frac{30 \times 0.15 \times 2}{30} N$
Molarity of KOH = 0.3M

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