QuestionApril 24, 2025

12. Determine whether a precipitate of PbBr_(2)(K_(sp)=4.67times 10^-6) forms when 175.0 mL of a 0.035 m Pb(NO_(3))_(2) solution is mixed with 145.0 mL of a 0.015 M KBr solution. A. a precipitate will form because Q_(sp)lt K_(sp) B. a precipitate will not form because Q_(sp)lt K_(sp) C. a precipitate will form because Q_(sp)gt K_(sp) D. a precipitate will not form because Q_(sp)gt K_(sp) E. a precipitate will form because Q_(sp)=K_(sp)

12. Determine whether a precipitate of PbBr_(2)(K_(sp)=4.67times 10^-6) forms when 175.0 mL of a 0.035 m Pb(NO_(3))_(2) solution is mixed with 145.0 mL of a 0.015 M KBr solution. A. a precipitate will form because Q_(sp)lt K_(sp) B. a precipitate will not form because Q_(sp)lt K_(sp) C. a precipitate will form because Q_(sp)gt K_(sp) D. a precipitate will not form because Q_(sp)gt K_(sp) E. a precipitate will form because Q_(sp)=K_(sp)
12. Determine whether a precipitate of PbBr_(2)(K_(sp)=4.67times 10^-6) forms when 175.0 mL of a 0.035 m Pb(NO_(3))_(2) solution is mixed with 145.0 mL of a 0.015 M KBr solution. A. a precipitate will form because Q_(sp)lt K_(sp) B. a precipitate will not form because Q_(sp)lt K_(sp) C. a precipitate will form because Q_(sp)gt K_(sp) D. a precipitate will not form because Q_(sp)gt K_(sp) E. a precipitate will form because Q_(sp)=K_(sp)

Solution
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Answer

B. a precipitate will not form because Q_{sp}\lt K_{sp} Explanation 1. Calculate final concentrations Use dilution formula C_1V_1 = C_2V_2. For Pb^{2+}: C_2 = \frac{0.035 \times 175.0}{320.0} = 0.0191 M. For Br^{-}: C_2 = \frac{0.015 \times 145.0}{320.0} = 0.0068 M. 2. Calculate Q_{sp} Q_{sp} = [Pb^{2+}][Br^{-}]^2 = (0.0191)(0.0068)^2 = 8.82 \times 10^{-7}. 3. Compare Q_{sp} with K_{sp} Q_{sp} = 8.82 \times 10^{-7} and K_{sp} = 4.67 \times 10^{-6}. Since Q_{sp} < K_{sp}, no precipitate forms.

Explanation

1. Calculate final concentrations<br /> Use dilution formula $C_1V_1 = C_2V_2$. For $Pb^{2+}$: $C_2 = \frac{0.035 \times 175.0}{320.0} = 0.0191$ M. For $Br^{-}$: $C_2 = \frac{0.015 \times 145.0}{320.0} = 0.0068$ M.<br /><br />2. Calculate $Q_{sp}$<br /> $Q_{sp} = [Pb^{2+}][Br^{-}]^2 = (0.0191)(0.0068)^2 = 8.82 \times 10^{-7}$.<br /><br />3. Compare $Q_{sp}$ with $K_{sp}$<br /> $Q_{sp} = 8.82 \times 10^{-7}$ and $K_{sp} = 4.67 \times 10^{-6}$. Since $Q_{sp} < K_{sp}$, no precipitate forms.
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