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

Question

Accepted Answer

# Explanation:
## Step1: Write the dissolution equation
$$\ce{Ag3PO4(s) <=> 3Ag+(aq) + PO4^{3-}(aq)}$$
Let the solubility of $\ce{Ag3PO4}$ be $s$. Then, $[\ce{Ag+}]=3s$ and $[\ce{PO4^{3 - }}]=s$

## Step2: Write the $K_{sp}$ expression
The solubility - product constant expression for $\ce{Ag3PO4}$ is $K_{sp}=[\ce{Ag+}]^{3}[\ce{PO4^{3 - }}]$

## Step3: Substitute the values of $[\ce{Ag+}]$ and $[\ce{PO4^{3 - }}]$
Given $s = 4.3\times10^{-5}\space M$, $[\ce{Ag+}]=3\times(4.3\times 10^{-5})\space M$ and $[\ce{PO4^{3 - }}]=4.3\times10^{-5}\space M$

$$
\begin{align*}
K_{sp}&=(3s)^{3}(s)\
&=27s^{4}
\end{align*}
$$

Substitute $s = 4.3\times10^{-5}$ into the equation:

$$
\begin{align*}
K_{sp}&=27\times(4.3\times 10^{-5})^{4}\
&=27\times(3.418801\times10^{-18})\
&=9.2307627\times 10^{-17}
\end{align*}
$$

# Answer:
$K_{sp}=9.2\times 10^{-17}$

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QUESTION IMAGE

Question

Explanation:

Step1: Write the dissolution equation

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

Step3: Substitute the values of \([\ce{Ag+}]\) and \([\ce{PO4^{3 - }}]\)

Answer: