Question

a student measures the oh⁻ concentration in a saturated aqueous solution of silver hydroxide to be 1.46×10⁻⁴ m. based on her data, the solubility product constant for silver hydroxide is

Explanation:

Step1: Write dissolution reaction

$\text{AgOH}(s)
ightleftharpoons \text{Ag}^+(aq) + \text{OH}^-(aq)$

Step2: Relate ion concentrations

From the reaction, $[\text{Ag}^+] = [\text{OH}^-] = 1.46 \times 10^{-4}\ \text{M}$

Step3: Calculate $K_{sp}$

$K_{sp} = [\text{Ag}^+][\text{OH}^-]$
$K_{sp} = (1.46 \times 10^{-4})(1.46 \times 10^{-4})$
$K_{sp} = (1.46)^2 \times 10^{-8}$
$K_{sp} = 2.1316 \times 10^{-8}$

Answer:

$2.13 \times 10^{-8}$ (rounded to three significant figures)