##”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”^-##?

The ##K_”b”## of ##”Z”^-## is ##2.1 × 10^-10##.
The equation for the equilibrium is:
##”Z”^(-) + “H”_2″O” ⇌ “HZ” + “OH”^-##
##[“Z”^-]_0 = “0.350 mol”/”1.0 L” = “0.35 mol/L”##
The ##K_”b”## expression is:
##K_”b” = ([“HZ”][“OH”^-])/([“Z”^-])##
##”[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##

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