QuestionJanuary 7, 2026

A solution contains 0.018 moles each of I',Br'' and Cl^- When the solution is mixed with 200 mL of 0.24MAgNO_(3), how much AgCl(s) precipitates out? KspAgI=1.5times 10^-16 Ksp AgBr=5.0times 10^-13 Ksp AgCl=1.6times 10^-10 2.6g 3.3 g 0.0 g 1.7 g 5.0 g

A solution contains 0.018 moles each of I',Br'' and Cl^- When the solution is mixed with 200 mL of 0.24MAgNO_(3), how much AgCl(s) precipitates out? KspAgI=1.5times 10^-16 Ksp AgBr=5.0times 10^-13 Ksp AgCl=1.6times 10^-10 2.6g 3.3 g 0.0 g 1.7 g 5.0 g
A solution contains 0.018 moles each of I',Br'' and Cl^- When the solution is mixed with 200 mL of 0.24MAgNO_(3), how much AgCl(s) precipitates out? KspAgI=1.5times 10^-16 Ksp AgBr=5.0times 10^-13 Ksp AgCl=1.6times 10^-10 2.6g 3.3 g 0.0 g 1.7 g 5.0 g

Solution
3.8(236 votes)

Answer

1.7 g Explanation 1. Calculate moles of AgNO_3 added 0.24\,M \times 0.200\,L = 0.048 moles Ag^+ 2. Determine precipitation order (lowest K_{sp} first) Order: AgI (K_{sp}=1.5\times10^{-16}), then AgBr, then AgCl 3. Precipitate AgI 0.018 moles I^- react with 0.018 moles Ag^+ to form AgI. Remaining Ag^+ = 0.048 - 0.018 = 0.030 moles. 4. Precipitate AgBr 0.018 moles Br^- react with 0.018 moles Ag^+ to form AgBr. Remaining Ag^+ = 0.030 - 0.018 = 0.012 moles. 5. Check if AgCl will precipitate [Ag^+] = \frac{0.012}{0.2} = 0.06\,M, [Cl^-] = \frac{0.018}{0.2} = 0.09\,M Q = [Ag^+][Cl^-] = 0.06 \times 0.09 = 0.0054 Since Q \gg K_{sp,AgCl}, all Ag^+ will react with Cl^- until one is exhausted. 6. Calculate moles of AgCl formed 0.012 moles Ag^+ left, so 0.012 moles AgCl can form. 7. Convert moles AgCl to grams Molar mass AgCl = 143.5\,g/mol 0.012 \times 143.5 = 1.722\,g

Explanation

1. Calculate moles of $AgNO_3$ added<br /> $0.24\,M \times 0.200\,L = 0.048$ moles $Ag^+$<br /><br />2. Determine precipitation order (lowest $K_{sp}$ first)<br /> Order: $AgI$ ($K_{sp}=1.5\times10^{-16}$), then $AgBr$, then $AgCl$<br /><br />3. Precipitate $AgI$<br /> $0.018$ moles $I^-$ react with $0.018$ moles $Ag^+$ to form $AgI$. Remaining $Ag^+ = 0.048 - 0.018 = 0.030$ moles.<br /><br />4. Precipitate $AgBr$<br /> $0.018$ moles $Br^-$ react with $0.018$ moles $Ag^+$ to form $AgBr$. Remaining $Ag^+ = 0.030 - 0.018 = 0.012$ moles.<br /><br />5. Check if $AgCl$ will precipitate<br /> $[Ag^+] = \frac{0.012}{0.2} = 0.06\,M$, $[Cl^-] = \frac{0.018}{0.2} = 0.09\,M$<br /> $Q = [Ag^+][Cl^-] = 0.06 \times 0.09 = 0.0054$<br /> Since $Q \gg K_{sp,AgCl}$, all $Ag^+$ will react with $Cl^-$ until one is exhausted.<br /><br />6. Calculate moles of $AgCl$ formed<br /> $0.012$ moles $Ag^+$ left, so $0.012$ moles $AgCl$ can form.<br /><br />7. Convert moles $AgCl$ to grams<br /> Molar mass $AgCl = 143.5\,g/mol$<br /> $0.012 \times 143.5 = 1.722\,g$
Click to rate:

Similar Questions