Question

Equilibrium constant of $T_{2}O$ ($T$ or ${}_1{H}^3$ is an isotope of ${}_1{H}^1$) and $H_{2}O$ are different at 298K. Let at 298K pure $T_{2}O$ has $pT$ (like $pH$) is $7.62$. The $pT$ of a solution prepared by adding $10mL$ of $0.2MTCl$ to $15mL$ of $0.25MNaOT$ is:

• A. $2-\log 7$
• B. $14+\log 7$
• C. $1.324-\log 7$
• D. $13.24+\log 7$

Answer 1

The correct option is D $13.24+\log 7$

$TCl+NaOT⟶NaCl+T_{2}O$

$10\times 0.2=2$ $15\times 0.25=3.75$ --------- initial

Since NaOT is in excess,

Remaining $NaOT=1.75$

$\left[O{T}^{-}\right]=\frac{1.75}{10+15}=\frac{1.75}{25}=7\times {10}^{-2}M$

$pOT=-\log \left[O{T}^{-}\right]$

$=-\log (7\times {10}^{-2})=2-\log 7$

$pT+pOT=7.62\times 2$

$pT=15.24-pOT$

$=15.24-(2-\log 7)$

$pT=13.24+\log 7$

option $D$ is correct