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Standard XII

Chemistry

pH the Power of H

Equilibrium c...

Question

Equilibrium constant of $T_{2} O$ (T or $_{1}^{3} H$ is an isotope of $_{1}^{1} H$ ) and $H_{2} O$ are different at 298 K. At 298 K, pure $T_{2} O$ has pT (like pH) of 7.62. The pT of a solution prepared by adding 10 mL of 0.2 M $T C l$ to 15 mL of 0.25 M $N a O T$ is:

2 - l o g 7

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14 + l o g 7

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13.24 - l o g 7

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13.24 + l o g 7

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Solution

The correct option is D $13.24 + l o g 7$
The given species will follow below equilibrium
$T_{2} O ⇆ 2 T^{+} + O T^{-}$
$T C l + N a O T \to N a C l + T_{2} O$
Initially we have
$T C l = 2 m m o l e s a n d N a O T = 3.75 m m o l e s$
After neutralization in 25 ml we are left with 1.75 mmoles of $N a O T$ so $[O T^{-}] = \frac{1.75 \times 10^{- 3}}{25 \times 10^{- 3}} = 7 \times 10^{- 2} M$
$p O T = 2 - l o g 7$ also we Know $p K_{T_{2} O} = 2 \times 7.62 = 15.34$
$∴ p T = 13.24 + l o g 7$

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Equilibrium constant of $T_{2} O$ (T or $_{1}^{3} H$ is an isotope of $_{1}^{1} H$ ) and $H_{2} O$ are different at 298 K. At 298 K, pure $T_{2} O$ has pT (like pH) of 7.62. The pT of a solution prepared by adding 10 mL of 0.2 M $T C l$ to 15 mL of 0.25 M $N a O T$ is: