#"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"^-#?

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Answer 1 · 2015-04-22T15:15:03

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#