Ionic EquilibriumHard

Question

The [OH-] in 100.0 ml of 0.016 M-HCl (aq) is :

Options

A.5 × 1012 M
B.3 × 10-10 M
C.6.25 × 10-13 M
D.2.0 × 10-9 M.

Solution

[H+] = 0.016 M
[H+] [OH-] = 10-14  ⇒ [OH-] = = 6.25 × 10-13 M

Create a free account to view solution

View Solution Free
Topic: Ionic Equilibrium·Practice all Ionic Equilibrium questions

More Ionic Equilibrium Questions

The solubility products of Mg(OH)2, Cd(OH)2, Al(OH)3 and Zn(OH)2 are 4 × 10–11, 8 × 10–6, 8.5 × 10–23 and 1.8 × 10–14, r...pH of 0.1M Na2HPO4 and 0.2M NaH2PO4 are respectively : pKa for H3PO4 are 2.2, 7.2 and 12.0....What is the pH of a neutral solution at 37°C, where Kw equals 2.5 × 10–14? (log 2 = 0.3)...Consider an aqueous solution, 0.1 M each in HOCN, HCOOH, (COOH)2 and H3PO4, for HOCN, we can write Ka(HOCN) = . [H+] in ...M(OH)x has KSP 4 × 10-12 and solubility 10-4 M. The value of x is:...