Calculate the mass of Ag2CO3 needed to form 1.00 L of saturated solution.

Please help! I have a huge test tomorrow. And can you please explain how you got the answer?

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To do this you need to specify the temperature. Since you have not specified I will assume it as room temperature which is 25 C.

Then you have to write down the equation for dissolving of Ag2CO3 in pure water. This is a reversible reaction.

the equation is,

Ag2CO3 (solid) <--------------> 2 Ag+ (aq) + CO3 (2-) (aq)

1 mole                                          2 moles       1 mole

aq - Aqueous

After this we have to find the solubility constant, in other words you have to find the equlibrium constant of this reaction.

Ksp = [Ag+]^2 x [CO3(2-)] = 8.46 x 10^(-12) ------- (1)

I have found this in internet, you can see the link below.

Now if you look at the above euqation, the Concentration of Ag+ is twice as that of Carbonate ions,


[Ag+]^2 = 2 x [CO3(2-)]


By substituting this in Equation (1), you get

8.46 x 10 ^(-12) = (2 x [CO3(2-)])^2 x [CO3(2-)]

8.46 x 10^(-12) = 4 x [CO3(2-)]^3

[CO3(2-)]^3 = 2.115 x 10^(-12)

[CO3(2-)] = 1.2836 x 10^(-04) mol per dm3


[CO3(2-)] = 1.2836 x 10^(-04) mol per litre

That means, in one liter of saturated solution of Ag2CO3, you have 1.2836 x 10^(-04) mol. To get this amount, from simple stochiometry, you need the exact same number of moles of Ag2CO3.

Therefore the mols of Ag2CO3 needed = 1.2836 x 10^(-04) mol

The Molecular weight of Ag2CO3,

Molecular weight of Ag 107.86, C - 12 and O - 16 respectively.

The Molecular weight of Ag2CO3 = 2*107.86 + 12 + 3*16

                                               = 275.72

Therefore the mass needed is  = 275.72 * 1.2836 x 10^(-04)

                                            = 0.03539 g.


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