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Question

Calculate whether or not a precipitate will form in the following mixed solutions:

(a) 100 mL100 \mathrm{~mL} of 0.010 M Na2SO40.010~M\mathrm{~Na}_2 \mathrm{SO}_4 and 100 mL100 \mathrm{~mL} of 0.001 M Pb(NO3)20.001~M\mathrm{~Pb}\left(\mathrm{NO}_3\right)_2
(b) 50.0 mL50.0 \mathrm{~mL} of 1.0×104 M AgNO31.0 \times 10^{-4}~M\mathrm{~AgNO}_3 and 100.mL100 . \mathrm{mL} of 1.0×104 M NaCl1.0 \times 10^{-4}~M\mathrm{~NaCl}
(c) 1.0 g Ca(NO3)21.0 \mathrm{~g} \mathrm{~Ca}\left(\mathrm{NO}_3\right)_2 in 150 mL H2O150 \mathrm{~mL} \mathrm{~H}_2 \mathrm{O} and 250 mL250 \mathrm{~mL} of 0.01 M NaOH0.01~M\mathrm{~NaOH}\

Ksp PbSO4=1.3×108K_{\text {sp }} \mathrm{PbSO}_4=1.3 \times 10^{-8}

KspAgCl=1.7×1010K_{\mathrm{sp}} \mathrm{AgCl}=1.7 \times 10^{-10}

KspCa(OH)2=1.3×106K_{\mathrm{sp}} \mathrm{Ca}(\mathrm{OH})_2=1.3 \times 10^{-6}

Solution

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Answered 2 years ago
Answered 2 years ago
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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 KspK_{sp}. If the product is less than the KspK_{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=cVn=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=nVc=\dfrac{n}{V}

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

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