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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" $5 脳 10^{- 8} M$ .

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:

$CuBr (s) 鈬� {Cu}^{+} + {Br}^{-}$

The

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

Therefore,

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

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Most popular questions from this chapter

Find $[{OH}^{-}]$ iflocalid="1663407699090" $[H^{+}] = 1.0 脳 10^{- 4} M [H^{+}] = 1.0 脳 10^{- 4} M$

Find $[{Cu}^{2 +}]$ in a solution saturated with ${Cu}_{4} {(OH)}_{6} ({so}_{4})$ if $[{OH}^{-}]$ is fixed at $1.0 脳 10^{- 6} M$ . Note that ${Cu}_{4} {(OH)}_{6} ({so}_{4})$ gives 1mol of ${SO}_{4}^{2 -}$ 4 mol for of ${Cu}^{2 +}$ .

${Cu}_{4} {(OH)}_{6} ({so}_{4}) (s) 鈬� 4 C u^{2 +} + 6 O H^{-} + S O_{4}^{2 -} K_{s p} = 2.3 脳 10^{- 69}$

Although $KOH, RbOH$ and CsOHhave little association between metal and hydroxide in aqueous solution, ${Li}^{+}$ and ${Na}^{+}$ do form complexes with:

role="math" localid="1663390733265" ${Li}^{+} + {OH}^{-} 鈬� LiOH (aq) K_{1} = \frac{[LiOH (aq)]}{[{Li}^{+}] [{OH}^{-}]} = 0.83$

${Na}^{+} + {OH}^{-} 鈬� NaOH (aq) K_{1} = 0.20$

Prepare a table like the one in Exercise 6-B showing initial and final Concentrations of ${Na}^{+} + {OH}^{-} 鈬� NaOH (aq)$ and (aq)insolution. Calculate the fraction of sodium in the form (aq)at equilibrium.

Solubility products predict that cation ${AX}^{3 +}$ can be 99.999% separated from cation $B^{2 +}$ by precipitation with anion X . When the separation is tried, we find 0.2% contamination of ${AX}_{3} (s)$ with $B^{2 +}$ . Explain what might be happening.

From the equations

$HOCl " 鈬� H^{+} + {OCl}^{-} K = 3.0 脳 10^{- 8} HOCl + {OBr}^{-} 鈬� HOBr + {OCl}^{-} K = 15$

find the value of K for the reaction $HOBr 鈬� H^{+} + {OBr}^{-}$ .

See all solutions

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