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Chapter 6: Q14 P (page 142)

Find $[Cu *]$ in equilibrium with CuBr(s) and 0.10MB $r^{-}$ .

Short Answer

Expert verified

The $[{Cu}^{+}]$ was calculated as role="math" localid="1663330959468" $5 \times 10^{- 8} M$ .

Step by step solution

01

The concept used.

Solubility Product:

The solubility product is a kind of equilibrium constant whose value depends on temperature.

02

The concentration

For given data, the equilibrium reaction is:

$CuBr (s) ⇌ {Cu}^{+} + {Br}^{-}$

The

$\begin{array}{l} [C u^{+}] [B r^{-}] = K_{s p} \\ [C u^{+}] [0.10] = 5 \times 10^{- 9} M \end{array}$

Therefore,

$\begin{array}{l} [{Cu}^{+}] = \frac{5 \times 10^{- 8} M}{0.10} \\ [{Cu}^{+}] = 5 \times 10^{- 8} M \end{array}$

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Most popular questions from this chapter

Calculate $[H^{+}]$ andpHfor the following solutions:

(a) $0.010 M {HNO}_{3}$

(b) $0.035 M KOH$

(c) $0.030 M HCI$

(d) $3.0 M HCI$

(e) $0.010 M [{({CH}_{3})}_{4} N^{+}] {OH}^{-}$

Tetramethylammonium hydroxide

Which will be more soluble (moles of metal dissolved per litre of solution), $B a {(I O_{3})}_{2} (K_{s p} = 1.5 \times 10^{- 9})$ or $C a {({IO}_{3})}_{2} (K_{sp} = 7.1 \times 10^{7})$ ? Give an example of a chemical reaction that might occur that would reverse the predicted solubilities.

Reaction 6-8 is allowed to come to equilibrium in a solution initially containing $0.0100 {MBrO}_{3}^{-}$ , $0.0100 {MCr}^{3 +}$ and 1.00MH⁺. To find the concentrations at equilibrium, we construct the table at the bottom of the page showing initial and final concentrations. We use the stoichiometry coefficients of the reaction to say that if $x m ol$ of ${Br}^{-}$ are created, then we must also make x mol of ${Cr}_{2} O_{7}^{2 -}$ and 8x mol of H⁺. To produce x mol of ${Br}^{-}$ , we must have consumed x mol of ${Br}^{-} O_{3}^{-}$ and 2x mol of Cr³⁺.

(a) Write the equilibrium constant expression that you would use to solve for x to find the concentrations at equilibrium. Do not try to solve the equation.

(b) Because $K = 1 \times 10^{11}$ , we suppose that the reaction will go nearly "to completion." That is, we expect both the concentration of ${Br}^{-}$ and ${Cr}_{2} O_{7}^{2 -}$ to be close to 0.00500M an equilibrium. (Why?) That is, $x \approx 0.00500 M$ . With this value of $x, [H^{+}] = 1.00 + 8 x = 1.04 M$ and $[{BrO}_{3}^{-}] = 0.0100 - x = 0.0050 M$ . However, we cannot say $[{Cr}^{3 +}] = 0.0100 - 2 x = 0$ , because there must be some small concentration of ${Cr}^{3 +}$ at equilibrium. Write $[{Cr}^{3 +}]$ for the concentration of ${Cr}^{3 +}$ and solve for $[{Cr}^{3 +}]$ . The limiting reagent in this example is ${Cr}^{3 +}$ . The reaction uses up before consuming ${BrO}_{3}^{-}$ .

Use Table 6-1 to find the $[H^{+}]$ in water at $100 ° C$ and at $0 ° C$ .

For the reaction ${HCO}_{3}^{-} ⇌ H^{+} + " {CO}_{3}^{2 -}$ , $ΔG ° = + 59.0 kJ / mol$ at $298.15 K$ . Find the value of Kfor the reaction.

See all solutions

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