For the reaction below, Kc = 1.10 × 10⁻⁸. What is the equilibrium concentration of OH⁻ if the reaction begins with 0.660 M HONH₂? HONH₂ (aq) + H₂O (l) ⇌ HONH₃⁺ (aq) + OH⁻ (aq)

Answer 1

Set up an ICE table

HONH₂(aq) + H₂O(l) <==> HONH₃⁺(aq) + OH^-(aq)

0.660.............................................0...................0................Initial

-x.................................................+x..................+x...............Change

0.660-x..........................................x....................x...............Equilibrium

Kc = 1.10x10^-8 = [HONH₃⁺][OH^-] / [HONH₂]

1.10x10^-8 = (x)(x) / 0.660 - x (since K is so small, ignore x in the denominator)

1.10x10^-8 = x² / 0.660

x² = 7.26x10^-9

x = 8.52x10^-5 M (above assumption was valid and subtracting this from 0.660 would be insignificant)

Equilibrium [OH^-] = 8.52x10^-5 M