a solution prepared in the lab contains $3.5 imes10^{-4}m sr^{2+}$ and $0.0010m f^{-}$. will a precipitate form? ($k_{sp}$ of $srf_{2}$ is $4.3 imes10^{-9}$)

Question

a solution prepared in the lab contains $3.5	imes10^{-4}m sr^{2+}$ and $0.0010m f^{-}$. will a precipitate form? ($k_{sp}$ of $srf_{2}$ is $4.3	imes10^{-9}$)

Accepted Answer

# Explanation:
## Step1: Write the solubility - product expression
For $SrF_{2}(s)
ightleftharpoons Sr^{2 +}(aq)+2F^{-}(aq)$, the solubility - product expression is $Q = [Sr^{2+}][F^{-}]^{2}$

## Step2: Substitute the given concentrations into the $Q$ expression
Given $[Sr^{2+}]=3.5\times 10^{-4}\space M$ and $[F^{-}]=0.0010 = 1.0\times 10^{-3}\space M$

$$
\begin{align*}
Q&=(3.5\times 10^{-4})\times(1.0\times 10^{-3})^{2}\
&=(3.5\times 10^{-4})\times(1.0\times 10^{-6})\
&=3.5\times 10^{-10}
\end{align*}
$$

## Step3: Compare $Q$ with $K_{sp}$
Given $K_{sp}=4.3\times 10^{-9}$

Since $Q = 3.5\times 10^{-10}$ and $K_{sp}=4.3\times 10^{-9}$, and $Q<K_{sp}$

# Answer:
No, a precipitate will not form.

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QUESTION IMAGE

Question

Explanation:

Step1: Write the solubility - product expression

Step2: Substitute the given concentrations into the \(Q\) expression

Step3: Compare \(Q\) with \(K_{sp}\)

Answer: