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Chapter 16: Problem 56

If \(10.0 \mathrm{mL}\) of $2.0 \times 10^{-3} M \mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}\( is added to 10.0 \)\mathrm{mL}$ of a \(\mathrm{pH}=10.0 \mathrm{NaOH}\) solution, will a precipitate form?

Short Answer

Expert verified
A precipitate will not form when 10.0 mL of \(2.0 \times 10^{-3} M\) Cr(NO₃)₃ is added to 10.0 mL of pH=10.0 NaOH solution, as the calculated solubility product (Q) is \(1.25 \times 10^{-15}\), which is less than the solubility product constant (Ksp) of Cr(OH)₃, approximately \(6.3 \times 10^{-32}\).

Step by step solution

01

Calculate the concentration of OH⁻ ions in NaOH solution

Given the pH of the solution, we can find the concentration of OH⁻ ions. pH = 10.0 pOH = 14 - pH = 14 - 10 = 4.0 The concentration of OH⁻ ions is: \[OH^-] = 10^{-pOH} = 10^{-4} = 1.0 \times 10^{-4} M\]
02

Find the initial concentration of Cr³⁺ and OH⁻ ions

When two solutions with equal volumes (10.0 mL each) are mixed, the total volume becomes 20.0 mL. We can now find the concentration of ions in the mixture: \[Cr^{3+}]_{initial} = \frac {2.0 \times 10^{-3} M \times 10.0 mL}{20.0 mL} = 1.0 \times 10^{-3} M\] \[OH^-]_{initial} = \frac {1.0 \times 10^{-4} M \times 10.0 mL}{20.0 mL} = 0.5 \times 10^{-4} M\]
03

Write the solubility product expression and find Q

The balanced chemical equation for the possible precipitation reaction is: \[Cr^{3+} + 3OH^- \longleftrightarrow Cr(OH)_3\] Now, let's write the solubility product expression and find the reaction quotient Q: \[Q = [Cr^{3+}][OH^-]^3\] Substitute the initial concentrations: \[Q = (1.0 \times 10^{-3})(0.5 \times 10^{-4})^3 = 1.25 \times 10^{-15}\]
04

Compare Q with Ksp to determine if precipitation would occur

For a precipitate to form, the value of Q must be greater than or equal to Ksp. The Ksp of Cr(OH)₃ is approximately 6.3 × 10⁻³². Comparing the values: \(Q = 1.25 \times 10^{-15}\) \(Ksp = 6.3 \times 10^{-32}\) Since \(Q > Ksp\), a precipitate will not form. Therefore, adding 10.0 mL of 2.0 × 10⁻³ M Cr(NO₃)₃ to 10.0 mL of pH=10.0 NaOH solution will not result in the formation of a precipitate.

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Most popular questions from this chapter

A solution contains $3.0 \times 10^{-3} M \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2} .$ What concentrations of \(\mathrm{KF}\) will cause precipitation of solid \(\mathrm{MgF}_{2}\left(K_{\mathrm{sp}}=6.4 \times 10^{-9}\right) ?\)

The \(K_{\mathrm{sp}}\) for lead iodide \(\left(\mathrm{PbI}_{2}\right)\) is $1.4 \times 10^{-8} .$ Calculate the solubility of lead iodide in each of the following. a. water b. \(0.10M\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) c. \(0.010 M\) \(\mathrm{NaI}\)

Calculate the mass of manganese hydroxide that dissolves to form 1300 mL of a saturated manganese hydroxide solution. For $\mathrm{Mn}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=2.0 \times 10^{-13}.$

For which salt in each of the following groups will the solubility depend on \(\mathrm{pH}\) ? $\begin{array}{ll}{\text { a. AgF, AgCl, AgBr }} & {\text { c. } \operatorname{Sr}\left(\mathrm{NO}_{3}\right)_{2}, \operatorname{Sr}\left(\mathrm{NO}_{2}\right)_{2}} \\ {\text { b. } \mathrm{Pb}(\mathrm{OH})_{2}, \mathrm{PbCl}_{2}} & {\text { d. } \mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{Ni}(\mathrm{CN})_{2}}\end{array}$

A solution is prepared by mixing 100.0 \(\mathrm{mL}\) of \(1.0 \times 10^{-4} M\) \(\mathrm{Be}\left(\mathrm{NO}_{3}\right)_{2}\) and 100.0 \(\mathrm{mL}\) of $8.0 M\( \)\mathrm{NaF}.$ $$\mathrm{Be}^{2+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BeF}^{+}(a q) \quad K_{1}=7.9 \times 10^{4}$$ $$\operatorname{Be} \mathrm{F}^{+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \operatorname{Be} \mathrm{F}_{2}(a q) \quad K_{2}=5.8 \times 10^{3}$$ $$\operatorname{BeF}_{2}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \operatorname{Be} \mathrm{F}_{3}^{-}(a q) \quad K_{3}=6.1 \times 10^{2}$$ $$\operatorname{Be} \mathrm{F}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BeF}_{4}^{2-}(a q) \qquad K_{4}=2.7 \times 10^{1}$$ Calculate the equilibrium concentrations of $\mathrm{F}^{-}, \mathrm{Be}^{2+}, \mathrm{BeF}^{+},\( \)\mathrm{BeF}_{2}, \mathrm{BeF}_{3}^{-},$ and \(\mathrm{BeF}_{4}^{2-}\) in this solution.
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