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Chapter 4: Problem 93

Suppose you have a solution that might contain any or all of the following cations: \(\mathrm{Ni}^{2+}, \mathrm{Ag}^{+}, \mathrm{Sr}^{2+},\) and \(\mathrm{Mn}^{2+}\). Addition of HCl solution causes a precipitate to form. After filtering off the precipitate, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution is added to the resulting solution and another precipitate forms. This is filtered off, and a solution of \(\mathrm{NaOH}\) is added to the resulting solution. No precipitate is observed. Which ions are present in each of the precipitates? Which of the four ions listed above must be absent from the original solution?

Short Answer

Expert verified
In the given solution, the first precipitate (with HCl) contains Ag鈦� as \(\mathrm{AgCl}\), and the second precipitate (with H鈧係O鈧�) contains Sr虏鈦� as \(\mathrm{SrSO}_4\). No precipitate is observed with NaOH, which means both Ni虏鈦� and Mn虏鈦� must be absent from the original solution.

Step by step solution

01

Determining precipitate with HCl

Addition of HCl solution would cause the following precipitate reactions: 1. \(\mathrm{Ag}^+ + \mathrm{Cl}^- \rightarrow \mathrm{AgCl}\) (precipitate) The other cations (Ni虏鈦�, Sr虏鈦�, and Mn虏鈦�) would not form a precipitate with HCl.
02

Determining precipitate with H鈧係O鈧�

After the precipitate from step 1 is filtered off, addition of H鈧係O鈧� solution would cause the following precipitate reactions: 1. \(\mathrm{Sr}^{2+} + \mathrm{SO}_4^{2-} \rightarrow \mathrm{SrSO}_4\) (precipitate) In this case, only Sr虏鈦� forms a precipitate with H鈧係O鈧�. The other two remaining cations (Ni虏鈦� and Mn虏鈦�) don't form a precipitate with H鈧係O鈧�.
03

Determining precipitate with NaOH

After the precipitate from step 2 is filtered off, addition of NaOH solution would cause the following reactions: 1. \(\mathrm{Ni}^{2+} + 2\mathrm{OH}^- \rightarrow \mathrm{Ni(OH)}_2\) (precipitate) 2. \(\mathrm{Mn}^{2+} + 2\mathrm{OH}^- \rightarrow \mathrm{Mn(OH)}_2\) (precipitate) In this step, both Ni虏鈦� and Mn虏鈦� would form precipitates with NaOH. However, it's observed that no precipitate forms when NaOH is added. This means neither Ni虏鈦� nor Mn虏鈦� are present in the solution at this step.
04

Conclusion

Based on the results from the three steps above, we can conclude the following: 1. The first precipitate (with HCl) contains Ag鈦� as \(\mathrm{AgCl}\) 2. The second precipitate (with H鈧係O鈧�) contains Sr虏鈦� as \(\mathrm{SrSO}_4\) 3. Since no precipitate is observed with NaOH, both Ni虏鈦� and Mn虏鈦� must be absent from the original solution.

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Most popular questions from this chapter

Suppose you have \(5.00 \mathrm{~g}\) of powdered magnesium metal, \(1.00 \mathrm{~L}\) of \(2.00 \mathrm{M}\) potassium nitrate solution, and \(1.00 \mathrm{~L}\) of \(2.00 \mathrm{M}\) silver nitrate solution. (a) Which one of the solutions will react with the magnesium powder? (b) What is the net ionic equation that describes this reaction? (c) What volume of solution is needed to completely react with the magnesium? (d) What is the molarity of the \(\mathrm{Mg}^{2+}\) ions in the resulting solution?

Some sulfuric acid is spilled on a lab bench. You can neutralize the acid by sprinkling sodium bicarbonate on it and then mopping up the resultant solution. The sodium bicarbonate reacts with sulfuric acid as follows: $$ \begin{array}{r} 2 \mathrm{NaHCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \\\ \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g) \end{array} $$ Sodium bicarbonate is added until the fizzing due to the formation of \(\mathrm{CO}_{2}(g)\) stops. If \(27 \mathrm{~mL}\) of \(6.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) was spilled, what is the minimum mass of \(\mathrm{NaHCO}_{3}\) that must be added to the spill to neutralize the acid?

(a) How many milliliters of a stock solution of \(6.0 \mathrm{M} \mathrm{HNO}_{3}\) would you have to use to prepare \(110 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{HNO}_{3} ?\) (b) If you dilute \(10.0 \mathrm{~mL}\) of the stock solution to a final volume of \(0.250 \mathrm{~L}\), what will be the concentration of the diluted solution?

The commercial production of nitric acid involves the following chemical reactions: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ (a) Which of these reactions are redox reactions? (b) In each redox reaction identify the element undergoing oxidation and the element undergoing reduction.

(a) Starting with solid sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\), describe how you would prepare \(250 \mathrm{~mL}\) of a \(0.250 \mathrm{M}\) sucrose solution. (b) Describe how you would prepare \(350.0 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) starting with \(3.00 \mathrm{~L}\) of \(1.50 \mathrm{M} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} .\)
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