Question #32388

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Question #32388

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Answer 1 · 2015-09-10T21:14:49

For part (b) $s = 1.2 \cdot 10^{- 4} M$ $s = 1.2 * 10^(-4)"M"$

Explanation:

I'll show you how to find the molar solubility of silver carboinate, ${Ag}_{2} {CO}_{3}$ $"Ag"_2"CO"_3$ .

Silver carbonate is considered insoluble in aqueous solution, which means that it does not dissociate completely into silver cations, ${Ag}^{2 +}$ $"Ag"^(2+)$ , and carbonate anions, ${CO}_{3}^{2 -}$ $"CO"_3^(2-)$ , when dissolved in water.

Actually, the fact that it doesn't dissociate completely is an understatement. When silver carbonate is placed in aqueous solution, an equilibrium reaction is established

${Ag}_{2} {CO}_{3(s]} ⇌ 2 {Ag}_{(aq]}^{+} + {CO}_{3(aq]}^{2 -}$ $"Ag"_2"CO"_text(3(s]) rightleftharpoons 2"Ag"_text((aq])^(+) + "CO"_text(3(aq])^(2-)$

The solubility product constant, $K_{s p}$ $K_(sp)$ , essentially tells you the extent of dissociation.

The smaller the value of $K_{s p}$ $K_(sp)$ , the fewer ions will dissociate in solution. Of course, this implies a lower solubility, since most of the solid will remain undissolved.

To find the molar solubility of silver carbonate, use an ICE table

${Ag}_{2} {CO}_{3(s]} ⇌ 2 {Ag}_{(aq]}^{+} + {CO}_{3(aq]}^{2 -}$ $"Ag"_2"CO"_text(3(s]) " "rightleftharpoons" " color(red)(2)"Ag"_text((aq])^(+)" " + " " "CO"_text(3(aq])^(2-)$

$I - 00$ $color(purple)("I")" " " " - " " " " " " " " " " " " 0 " " " " " " " " " " 0$
$C - (+ 2 s) (+ s)$ $color(purple)("C")" " " " - " " " " " " " " " " (+color(red)(2)s) " " " " " " "(+s)$
$E - 2 s s$ $color(purple)("E")" " " " - " " " " " " " " " " " " color(red)(2)s " " " " " " " " " " s$

By definition, $K_{s p}$ $K_(sp)$ will be

$K_{s p} = {[{Ag}^{+}]}^{2} \cdot [{CO}_{3}^{2 -}]$ $K_(sp) = ["Ag"^(+)]^color(red)(2) * ["CO"_3^(2-)]$

$K_{s p} = {(2 s)}^{2} \cdot s = 4 s^{2} \cdot s = 4 s^{3}$ $K_(sp) = (2s)^2 * s = 4s^2 * s = 4s^3$

This means that $s$ $s$ , which is the molar solubility of silver chloride, is equal to

$s = \sqrt[3]{\frac{K_{s p}}{4}} = \sqrt[3]{\frac{6.2 \cdot 10^{- 12}}{4}} = 1.2 \cdot 10^{- 4} M$ $s = root(3)(K_(sp)/4) = root(3)( (6.2 * 10^(-12))/4) = color(green)(1.2 * 10^(-4)"M")$

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Question #32388

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