Question

`"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"^-`?

Answer 1

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`