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Chapter 15: Problem 47

What mass of ZnS \(\left(K_{\mathrm{sp}}=2.5 \times 10^{-22}\right)\) will dissolve in \(300.0 \mathrm{mL}\) of \(0.050 \mathrm{M} \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2} ?\) Ignore the basic properties of \(\mathrm{S}^{2-}\).

Short Answer

Expert verified
The mass of ZnS that will dissolve in \(300.0 \mathrm{mL}\) of \(0.050 \mathrm{M} \mathrm{Zn}(\mathrm{NO}_{3})_{2}\) is approximately \(1.5 \times 10^{-19} \mathrm{g}\).

Step by step solution

01

Write the Ksp expression and replace the given values

For the reaction ZnS <=> Zn2+ + S2-, the Ksp expression is given by: Ksp = [Zn2+][S^2-] We are given Ksp = 2.5 × 10^-22 and [Zn2+] = 0.050 M, so we have: 2.5 × 10^-22 = (0.050)[S^2-]
02

Solve for the concentration of S^2- ions

Now we need to find the concentration of S^2-, so we simply solve for [S^2-]: [S^2-] = (2.5 × 10^-22) / (0.050) [S^2-] = 5.0 × 10^-21 M
03

Calculate the amount of moles of S^2- ions dissolving in the solution

To find the amount of moles of S^2- ions, we will use the formula: moles(S^2-) = [S^2-] × volume of solution (in L) We are given volume as 300.0 mL, which is equal to 0.300 L. So, moles(S^2-) = (5.0 × 10^-21 M)(0.300 L) moles(S^2-) = 1.5 × 10^-21 mol
04

Calculate the mass of ZnS dissolved in solution

To find the mass of ZnS that will dissolve in the solution, we first need the molar mass of ZnS. Molecular weight (ZnS) = atomic weight (Zn) + atomic weight (S) = 65.38 g/mol + 32.06 g/mol = 97.44 g/mol Now, we multiply the moles of S^2- ions (which will dissolve one mole of ZnS) by the molar mass of ZnS to find the mass: mass(ZnS) = moles(S^2-) × molar mass (ZnS) mass(ZnS) = (1.5 × 10^-21 mol)(97.44 g/mol) mass(ZnS) = approximately 1.5 × 10^-19 g So, the mass of ZnS that will dissolve in 300.0 mL of 0.050 M Zn(NO3)2 is approximately 1.5 × 10^-19 g.

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Most popular questions from this chapter

Calculate the solubility of solid \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\left(K_{\mathrm{sp}}=1.3 \times 10^{-32}\right)\) in a \(0.20-M \mathrm{Na}_{3} \mathrm{PO}_{4}\) solution.

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties. a. \(\mathrm{PbI}_{2}, K_{\mathrm{sp}}=1.4 \times 10^{-8}\) b. \(\mathrm{CdCO}_{3}, K_{\mathrm{sp}}=5.2 \times 10^{-12}\) c. \(\operatorname{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}, K_{\mathrm{sp}}=1 \times 10^{-31}\)

A solution contains \(0.25\) \(M\) \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}\) and \(0.25 M \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) Can the metal ions be separated by slowly adding \(\mathrm{Na}_{2} \mathrm{CO}_{3} ?\) Assume that for successful separation \(99 \%\) of the metal ion must be precipitated before the other metal ion begins to precipitate, and assume no volume change on addition of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\).

Consider \(1.0 \mathrm{L}\) of an aqueous solution that contains \(0.10\) \(M\) sulfuric acid to which 0.30 mole of barium nitrate is added. Assuming no change in volume of the solution, determine the \(\mathrm{pH},\) the concentration of barium ions in the final solution, and the mass of solid formed.

A solution is \(1 \times 10^{-4} M\) in \(\mathrm{NaF}, \mathrm{Na}_{2} \mathrm{S},\) and \(\mathrm{Na}_{3} \mathrm{PO}_{4} .\) What would be the order of precipitation as a source of \(\mathrm{Pb}^{2+}\) is added gradually to the solution? The relevant \(K_{\mathrm{sp}}\) values are \(K_{\mathrm{sp}}\left(\mathrm{PbF}_{2}\right)\) \(=4 \times 10^{-8}, K_{\mathrm{sp}}(\mathrm{PbS})=7 \times 10^{-29},\) and \(K_{\mathrm{sp}}\left[\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right]=\) \(1 \times 10^{-54}\).
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