# "Z"^- is a weak base. An aqueous solution prepared by dissolving 0.350 mol of "NaZ" in sufficient water to yield 1.0 L of solution has a pH of 8.93 at 25.0 °C. What is the K_"b" of "Z"^-?

Apr 22, 2015

The ${K}_{\text{b}}$ of ${\text{Z}}^{-}$ is 2.1 × 10^-10.

The equation for the equilibrium is:

${\text{Z"^(-) + "H"_2"O" ⇌ "HZ" + "OH}}^{-}$

["Z"^-]_0 = "0.350 mol"/"1.0 L" = "0.35 mol/L"

The ${K}_{\text{b}}$ expression is:

K_"b" = (["HZ"]["OH"^-])/(["Z"^-])

$\text{[pH](http://socratic.org/chemistry/acids-and-bases/the-ph-concept)} = 8.93$

"pOH" = 14.00 – 8.93 = 5.07

["OH"^-] = 10^"-pOH" = 10^-5.07 = 8.51 × 10^-6"mol/L"

["HZ"]= 8.51 × 10^-6"mol/L"

["Z"^-] = (0.35 - 8.51 × 10^-6) " mol/L" = "0.35 mol/L"

K_"b" = (["HZ"]["OH"^-])/(["Z"^-]) = (8.51 × 10^-6 × 8.51 × 10^-6)/0.35 = 2.1 × 10^-10