$"Z"^-$ 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" = 8.93#

$"pOH" = 14.00 – 8.93 = 5.07$

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

$["HZ"]= 8.51 × 10^-6"mol/L"$

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

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

"Z"^-"Z"^- is a weak base. An aqueous solution prepared by dissolving 0.350 mol of "NaZ""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"K_"b" of "Z"^-"Z"^-?