Question

what is the solubility, in moles per dm3 of co3^= ions in a saturated solution of silver carbonate Ag2Co2
a)1.15x10^-4
b)1.15x10^-6
c)2.48x10^-4
d)2.48x10^-6

Answer 1

Answer on Question #70646 – Chemistry – Other

Task:

What is the solubility, in moles per dm³ of CO₃²⁻ ions in a saturated solution of silver carbonate Ag₂CO₃.

A) 1.15*10⁻⁴;

B) 1.15*10⁻⁶;

C) 2.48*10⁻⁴;

D) 2.48*10⁻⁶.

Solution:

The dissociation equation of silver carbonate:

$$\begin{array}{c c} A g _ {2} C O _ {3} & 2 A g ^ {+} + C O _ {3} ^ {2 -} \\ 2 s & s \end{array}$$ $$[ A g ^ {+} ] = 2 s;$$ $$[ C O _ {3} ^ {2 -} ] = s;$$

The expression for the solubility constant of silver carbonate:

$$\mathrm {K} _ {\mathrm {s}} \left(A g _ {2} C O _ {3}\right) = \left[ A g ^ {+} \right] ^ {2} * \left[ C O _ {3} ^ {2 -} \right];$$ $$\mathrm {K} _ {\mathrm {s}} \left(A g _ {2} C O _ {3}\right) = 6. 1 5 * 1 0 ^ {- 1 2}.$$

Then,

$$\begin{array}{l} \mathrm {K} _ {\mathrm {s}} \left(A g _ {2} C O _ {3}\right) = \left[ A g ^ {+} \right] ^ {2} * \left[ C O _ {3} ^ {2 -} \right] = (2 s) ^ {2} * s = 4 s ^ {3} = 6. 1 5 * 1 0 ^ {- 1 2}. \\ s ^ {3} = \frac {6 . 1 5 * 1 0 ^ {- 1 2}}{4} = 1. 5 3 7 5 * 1 0 ^ {- 1 2}; \\ s = \sqrt [ 3 ]{1.5375 * 10 ^ {- 1 2}} = 1.15 * 1 0 ^ {- 4} \left(\frac {m o l}{d m ^ {3}}\right). \\ \end{array}$$

Answer: A) 1.15*10⁻⁴ mol/dm³.

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