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Math  /  Data & Statistics

QuestionWhen 22.0 mL of a 2.45×10−4 M sodium phosphate solution is combined with 15.0 mL of a 3.54×10−4 M chromium(III) acetate solution, does a precipitate form?\text{When } \mathbf{22.0 \, \text{mL}} \text{ of a } 2.45 \times 10^{-4} \, \text{M} \text{ sodium phosphate solution is combined with } 15.0 \, \text{mL} \text{ of a } 3.54 \times 10^{-4} \, \text{M} \text{ chromium(III) acetate solution, does a precipitate form?} \text{(yes or no)}
For these conditions the Reaction Quotient, Q, is equal to\text{For these conditions the Reaction Quotient, Q, is equal to}
\text{Hello! It looks like you're working on a chemistry problem involving precipitation and the Reaction Quotient (Q). To determine whether a precipitate forms, we need a bit more information, specifically the solubility product constant (KspK_{sp}) for the possible precipitate that might form from mixing these solutions.}
\text{Could you provide the KspK_{sp} value for the compound that might precipitate, or let me know which compound you suspect will precipitate? Once we have that, I can help you calculate Q and determine if a precipitate will form!}
\text{Not given}

Studdy Solution

STEP 1

1. The possible precipitate formed is chromium(III) phosphate, CrPO4\text{CrPO}_4.
2. The solubility product constant, KspK_{sp}, for CrPO4\text{CrPO}_4 is required to determine if a precipitate forms.
3. The reaction quotient, QQ, needs to be calculated and compared to KspK_{sp}.

STEP 2

1. Write the balanced chemical equation for the potential precipitation reaction.
2. Calculate the concentrations of ions after mixing.
3. Calculate the reaction quotient, QQ.
4. Compare QQ to KspK_{sp} to determine if a precipitate forms.

STEP 3

Write the balanced chemical equation for the potential precipitation reaction:
Cr3++PO43−→CrPO4(s)\text{Cr}^{3+} + \text{PO}_4^{3-} \rightarrow \text{CrPO}_4(s)

STEP 4

Calculate the total volume after mixing:
22.0 mL+15.0 mL=37.0 mL22.0 \, \text{mL} + 15.0 \, \text{mL} = 37.0 \, \text{mL}
Calculate the concentration of Cr3+\text{Cr}^{3+} ions after mixing:
[Cr3+]=(15.0 mL×3.54×10−4 M37.0 mL)=1.43×10−4 M[\text{Cr}^{3+}] = \left( \frac{15.0 \, \text{mL} \times 3.54 \times 10^{-4} \, \text{M}}{37.0 \, \text{mL}} \right) = 1.43 \times 10^{-4} \, \text{M}
Calculate the concentration of PO43−\text{PO}_4^{3-} ions after mixing:
[PO43−]=(22.0 mL×2.45×10−4 M37.0 mL)=1.46×10−4 M[\text{PO}_4^{3-}] = \left( \frac{22.0 \, \text{mL} \times 2.45 \times 10^{-4} \, \text{M}}{37.0 \, \text{mL}} \right) = 1.46 \times 10^{-4} \, \text{M}

STEP 5

Calculate the reaction quotient, QQ:
Q=[Cr3+][PO43−]=(1.43×10−4 M)(1.46×10−4 M)=2.09×10−8Q = [\text{Cr}^{3+}][\text{PO}_4^{3-}] = (1.43 \times 10^{-4} \, \text{M})(1.46 \times 10^{-4} \, \text{M}) = 2.09 \times 10^{-8}

STEP 6

Without the KspK_{sp} value for CrPO4\text{CrPO}_4, we cannot definitively determine if a precipitate forms. However, if Q>KspQ > K_{sp}, a precipitate will form; if Q<KspQ < K_{sp}, no precipitate will form.
Since the KspK_{sp} value is not provided, we cannot conclude whether a precipitate forms. Please provide the KspK_{sp} value for a complete solution.

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