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# Calculate whether or not a precipitate will form in the following mixed solutions:(a) $100 \mathrm{~mL}$ of $0.010~M\mathrm{~Na}_2 \mathrm{SO}_4$ and $100 \mathrm{~mL}$ of $0.001~M\mathrm{~Pb}\left(\mathrm{NO}_3\right)_2$ (b) $50.0 \mathrm{~mL}$ of $1.0 \times 10^{-4}~M\mathrm{~AgNO}_3$ and $100 . \mathrm{mL}$ of $1.0 \times 10^{-4}~M\mathrm{~NaCl}$ (c) $1.0 \mathrm{~g} \mathrm{~Ca}\left(\mathrm{NO}_3\right)_2$ in $150 \mathrm{~mL} \mathrm{~H}_2 \mathrm{O}$ and $250 \mathrm{~mL}$ of $0.01~M\mathrm{~NaOH}$\$K_{\text {sp }} \mathrm{PbSO}_4=1.3 \times 10^{-8}$$K_{\mathrm{sp}} \mathrm{AgCl}=1.7 \times 10^{-10}$$K_{\mathrm{sp}} \mathrm{Ca}(\mathrm{OH})_2=1.3 \times 10^{-6}$

Solution

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In this task, we need to determine if precipitate will form.

The precipitate will form when the the product of the molar concentration of the ions in solution ( raised to the potency of its stoichiometric factor) is greater than the $K_{sp}$. If the product is less than the $K_{sp}$, the precipitate will not occur.

From the dissociation reaction, we determine the mole ratio of ion and solution

We use the molarity and volume of the solution to determine how many moles of ion we have:

$n=c \cdot V$

Then we calculate total volume of solution that is the sum of volume of two solution. Then, we calculate concentration of each ion using the moles and total volume:

$c=\dfrac{n}{V}$

Then we write the reaction quotient ($Q_{sp}$) using the initial concentration of ions. If the $Q_{sp}$>$K_{sp}$, the precipitate will occur.

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