Question #17cc3

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Question #17cc3

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Answer 1 · 2015-04-13T10:13:33

The answer is A) $2 \cdot 10^{- 3} M$ $2 * 10^(-3)"M"$ .

Because the solubility product constant, $K_{s p}$ $K_(sp)$ , of manganese (II) hydroxide is so small, the compound is considered insoluble in aqueous solution.

However, very, very little amounts will dissociate according to the balanced equilibrium equation

$M n {(O H)}_{2(s]} ⇌ M n_{(aq]}^{2 +} + 2 O H_{(aq]}^{-}$ $Mn(OH)_text(2(s]) rightleftharpoons Mn_text((aq])^(2+) + color(red)(2)OH_text((aq])^(-)$

According to the definition of the solutbility constant product, you have

$K_{s p} = [M n^{2 +}] \cdot {[O H^{-}]}^{2}$ $K_(sp) = [Mn^(2+)] * [OH^(-)]^(color(red)(2))$ $(1)$ $color(blue)((1))$

Every time the concentrations of the $M n^{2 +}$ $Mn^(2+)$ and $O H^{-}$ $OH^(-)$ ions satisfy the above equation, a precipitate will form.

You can determine the concentration of the hydroxide ions from the solution's pH

$p O H = 14 - p H_{sol} = 14 - 9 = 5$ $pOH = 14 - pH_"sol" = 14 - 9 = 5$

$[O H^{-}] = 10^{- p O H} = 10^{- 5}$ $[OH^(-)] = 10^(-pOH) = 10^(-5)$

Therefore, according to equation $(1)$ $color(blue)((1))$ ,

$[M n^{2 +}] = \frac{K_{s p}}{{([O H^{-}])}^{2}} = \frac{2 \cdot 10^{- 13}}{{(10^{- 5})}^{2}} = \frac{2 \cdot 10^{- 13}}{10^{- 10}}$ $[Mn^(2+)] = K_(sp)/([OH^(-)])^(2) = (2 * 10^(-13))/(10^(-5))^(2) = (2 * 10^(-13))/10^(-10)$

$[M n^{2 +}] = 2 \cdot 10^{- 3} M$ $[Mn^(2+)] = color(green)(2 * 10^(-3)"M")$

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Question #17cc3

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