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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 T2O (T or 31H is an isotope of 11H) and H2O are different at 298 K. At 298 K, pure T2O has pT (like pH) of 7.62. The pT of a solution prepared by adding 10 mL of 0.2 M TCl to 15 mL of 0.25 M NaOT is:

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: