Question

a student measures the molar solubility of silver carbonate in a water solution to be 1.29×10⁻⁴ m. based on her data, the solubility product constant for this compound is an error has been detected in your answer. check for typos, miscalculations etc. before submitting your answer submit answer retry entire group 9 more group attempts remaining

Explanation:

Step1: Write dissolution reaction

$\text{Ag}_2\text{CO}_3(s)
ightleftharpoons 2\text{Ag}^+(aq) + \text{CO}_3^{2-}(aq)$

Step2: Define molar solubility

Let $s = 1.29 \times 10^{-4}\ \text{M}$. Then $[\text{Ag}^+] = 2s$, $[\text{CO}_3^{2-}] = s$.

Step3: Write $K_{sp}$ expression

$K_{sp} = [\text{Ag}^+]^2 [\text{CO}_3^{2-}]$

Step4: Substitute values into formula

$K_{sp} = (2s)^2 \times s = 4s^3$
$K_{sp} = 4 \times (1.29 \times 10^{-4})^3$

Step5: Calculate the result

First compute $(1.29 \times 10^{-4})^3 = 1.29^3 \times 10^{-12} \approx 2.1466 \times 10^{-12}$
Then $K_{sp} = 4 \times 2.1466 \times 10^{-12} \approx 8.59 \times 10^{-12}$

Answer:

$8.59 \times 10^{-12}$