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 $$${\text{Now, }P^{Kw}\text{ of }D_{2}O = P^{D} + P^{OD} = 13.6 + 1.4 = 15 }{\therefore Kw = 1 \times 10^{- 15}}$$
Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
pH of 0.01 M-(NH4)2SO4 and 0.02 M-NH4OH buffer (pKa of NH4+ = 9.26) is...An acid type indicator, HIn differs in colour from its conjugate base (In–). The human eye is sensitive to colour differ...An acid solution with pH = 6 at 25oC is diluted by 102 times. The pH of solution will:...How many times the solubility of CaF2 is decreased in 4 × 10−3 M-KF(aq) solution as compared to pure water at 25o C. Giv...Which of the following solution will have a pH exactly equal to 8 ?...