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 \(s\). From the equation, \([\ce{Ag+}]=3s\) and \([\ce{PO4^{3 - }}]=s\).

Step2: Write the \(K_{sp}\) expression

The solubility - product constant expression for \(\ce{Ag3PO4}\) is \(K_{sp}=[\ce{Ag+}]^{3}[\ce{PO4^{3 - }}]\).
Substitute \([\ce{Ag+}]=3s\) and \([\ce{PO4^{3 - }}]=s\) into the \(K_{sp}\) expression:
\(K_{sp}=(3s)^{3}\times s\).
Simplify the expression: \(K_{sp}=27s^{4}\).

Step3: Substitute the value of \(s\)

Given \(s = 4.3\times10^{-5}\space M\).
\(K_{sp}=27\times(4.3\times 10^{-5})^{4}\)
First, calculate \((4.3\times 10^{-5})^{4}=4.3^{4}\times10^{-20}\)
\(4.3^{4}=(4 + 0.3)^{4}=4^{4}+4\times4^{3}\times0.3+6\times4^{2}\times0.3^{2}+4\times4\times0.3^{3}+0.3^{4}=256+192 + 8.64+0.432 + 0.0081=457.0801\)
\(K_{sp}=27\times457.0801\times10^{-20}\)
\(K_{sp}=1.23411627\times10^{-16}\approx1.2\times 10^{-16}\)

Answer:

\(K_{sp}\) for \(\ce{Ag3PO4}\) is \(1.2\times 10^{-16}\)