Question

the solubility of ag3po4 in water at 25 °c is 4.3 × 10^-5 m.
what is the k_sp for ag3po4?

Explanation:

Step1: Write the dissolution equation

$$\ce{Ag3PO4(s) <=> 3Ag+(aq) + PO4^{3-}(aq)}$$
Let the solubility of $\ce{Ag3PO4}$ be $s$. Then $[\ce{PO4^{3-}}]=s$ and $[\ce{Ag+}]=3s$.

Step2: Write the expression for $K_{sp}$

$$K_{sp}=[\ce{Ag+}]^3[\ce{PO4^{3-}}]$$
Substitute $[\ce{Ag+}]=3s$ and $[\ce{PO4^{3-}}]=s$ into the equation:
$$K_{sp}=(3s)^3\times s$$
$$K_{sp}=27s^4$$

Step3: Substitute the value of $s$

Given $s = 4.3\times10^{-5}\text{ M}$
$$K_{sp}=27\times(4.3\times 10^{-5})^4$$
First calculate $(4.3\times 10^{-5})^4=(4.3)^4\times(10^{-5})^4 = 341.8801\times10^{-20}$
Then $K_{sp}=27\times341.8801\times10^{-20}$
$$K_{sp}=9.2307627\times10^{-18}$$

Answer:

$9.2\times 10^{-18}$