When 20 ml of 0.2 M - DCl solution is mixed with 80 ml of 0.1 M - NaOD solution, pD of the resulting solution becomes 13.6. The ionic product of heavy water, , is

Question

When 20 ml of 0.2 M - DCl solution is mixed with 80 ml of 0.1 M - NaOD solution, pD of the resulting solution becomes 13.6. The ionic product of heavy water, $D_2O$, is

Prashant Kotian · Accepted Answer

Correct answer: A. Moles of D+ added = 20 ml × 0.2 M = 4.0 mmol.
Moles of OD- added = 80 ml × 0.1 M = 8.0 mmol.
Reaction: D+ + OD- -> .
Excess OD- = 8.0 - 4.0 = 4.0 mmol.
Total volume = 100 ml = 0.1 L.
Concentration of OD- in the resulting solution [OD-] = 4.0 mmol / 100 ml = 0.04 M = 4.0 × 10^-2 M.
Given resulting pD = 13.6, which means [D+] = 10^-13.6 M.
The ionic product of heavy water is KD2O = [D+][OD-] = 10^-13.6 × 4.0 × 10^-2 = (2.51 × 10^-14) × 4.0 × 10^-2 = 1.0 × 10^-15.

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