What is the molar solubility of AgI in 0.20 M NaCN?
(a) 6.2 x 10^-4 M
(b) 1.0 x 10^-1 M
(c) 7.6 x 10^-2 M
(d) 2.1 x 10^-3 M

Question

What is the molar solubility of AgI in 0.20 M NaCN? 
(a) 6.2 x 10^-4 M
(b) 1.0 x 10^-1 M
(c) 7.6 x 10^-2 M
(d) 2.1 x 10^-3 M

Accepted Answer

Hey everyone, we're asked to calculate the scalability of silver chloride in moles, per liters in 0.15 molar potassium theo sign it. And were provided are solid ability product constant of silver chloride and the molo freezing point depression constant for silver tetra thigh assign it first. Let's go ahead and determine the equilibrium for silver chloride. So for silver chloride We're going to end up with silver with a plus one charge plus chlorine with a -1 charge. Now our cell viability product constant for this is going to be the concentration of our silver ion times the concentration of our chlorine ion. And silver chloride is not included in this since it is a solid. Now let's go ahead and look at the formation of the complex of silver tetra assign it now to form this, we would have our silver plus our theo sign it. And to balance this out, we would need to add a coefficient of four prior to assassinate. And this is going to get us to the formation of silver to sign it. Now the mole freezing point depression constant for this is going to be the concentration of our silver tetra cosign it divided by the concentration of our silver ion times the concentration of our thio Sinai on. And this is going to be raised to the power of four due to that coefficient of four. Now let's go ahead and add these two equations. So our silver ion is going to cancel out and we're going to end up with silver chloride Plus four of our tae assign ion and our products are going to be our chlorine ion Plus Our Silver Tetra Dioxin eight. Now let's go ahead and create our chart. So our silver chloride won't be included in this since it is a solid. And we were told that we had 0.15 moller starting of our potassium niacin. It we had zero of our products formed initially and our change is going to be a minus four X for our diocesan ion. Since we have that coalition of four and a plus X on our product side since we're gaining products. Now when we bring this down we get 0.15 minus four X. And we get an X. And an X in our product side. Now our molo freezing point depression constant times our solid ability product constant is going to be equal to our products over our reactant. So we have our X times R. X. And this is going to be divided by 0.15 minus four X. Raised to the power of four due to that coefficient of four. Now let's go ahead and plug in our values. We were told that our molo freezing point depression constant was 1. times 10 to the 10. And this is going to be multiplied by our solid ability product constant which was said to be 1.8 times 10 to the -10. This will be equal to X squared Divided by 0.15 -4 X raised to the power of four. Now if we take the square root of both sides, We end up with 1.46969 equals x over 0.15 -4 x squared. Now let's go ahead and isolate our X. So we get 1.46969 time 0.15 minus four X squared, which is going to be equal to X. Now when we multiply this out, we got 1.4696, 9 times 0.225 plus 16 X squared minus 1.2 X. And this is going to be equal to X. Multiplying this out again, we get 0.3307 plus 23.515 X squared minus 1.763628 X. And again this is equal to X. Now let's go ahead and move our X to the left so we can make this equal to zero. So we get 0.03307 plus 23.515, X squared minus 2.763628, X equals zero, solving the quadratic equation. We get X equals 0.104 and x equals 0. 1352. Now we know that X cannot be 0.104. And the reason why is because if we plugged in 0.15 minus four times 0.104, We end up with negative 0. and we cannot have a negative polarity for the assigning ion in our cumulative equation. So the answer must be X equals 0.01352. So to answer this question and to further simplify it, The Mueller scalability of silver chloride is going to be 0.014 moller and this is going to be our final answer. Now, I hope that made sense and let us know if you have any questions.

What is the molar solubility of AgI in 0.20 M NaCN? (a) 6.2 x 10^-4 M (b) 1.0 x 10^-1 M (c) 7.6 x 10^-2 M (d) 2.1 x 10^-3 M

Verified Solution

Key Concepts

Molar Solubility

Common Ion Effect

Equilibrium Constant (Ksp)