Ionic EquilibriumHard
Question
When 20 ml of 0.2 M-DCl solution is mixed with 80 ml of 0.1 M-NaOD solution, pH of the resulting solution becomes 13.6. The ionic product of heavy water D2O is
Options
A.10−15
B.10−16
C.4 × 10−15
D.4 × 10−16
Solution
$\left\lbrack OD^{-} \right\rbrack_{\text{excess}} = \frac{80 \times 0.1 - 20 \times 0.2}{100} = 0.04\text{ M}$
$\therefore P^{OD} = - \log(0.04) = 1.4 $$${\text{Now, }P^{Kw}\text{ of }D_{2}O = P^{D} + P^{OD} = 13.6 + 1.4 = 15 }{\therefore Kw = 1 \times 10^{- 15}}$$
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