Question

At 25 degrees C the Ksp for SrSO4 is 7.6*10^-7 . The Ksp for SrF2 is 7.9*10^-10 .

a.) What is the molar solubility of SrSO4 in pure water?
b.) What is the molar solubilty of SrF2 in pure water?
C.) A solution of Sr(NO3)2 is added slowly to 1 L of a solution containing 0.020 mole F and 0.10 mole of SO4^2 Which salt precipitates first? What is the concentration of Sr^2 in the solution when the first precipitate begins to form?
D.) As more Sr(NO3)2 is added to the mixture in (c) a second precipitates begins to form. At that stage, what percent of the anion of the first precipitate remains in the solution?

Answer 1

The molar solubility of SrSO4 at 25°C in pure water is 8.71x10^-4 M and for SrF2 it is 6.45x10^-4 M. SrSO4 will precipitate first when Sr(NO3)2 is added to a mixture with fluoride and sulfate ions. The concentration of Sr^2+ will be approximately 3.8x10^-5 M at the onset of the first precipitation.

Ksp (Solubility Product Constant) calculations allow us to determine the molar solubility of sparingly soluble compounds. To find the molar solubility of SrSO4 in pure water, we set up the expression based on the equation SrSO4(s) ⇌ Sr2+(aq) + SO42-(aq), meaning Ksp = [Sr2+][SO42-] = s2, where s is the solubility. Plugging in Ksp = 7.6x10-7, we find s = √(7.6x10-7) and conclude that s = 8.71x10-4 M. For SrF2, we use SrF2(s) ⇌ Sr2+(aq) + 2F-(aq), getting Ksp = [Sr2+][F-]2 = s(2s)2 = 4s3. Substituting Ksp = 7.9x10-10, we solve for s, yielding s = 6.45x10-4 M.

Regarding which salt precipitates first when Sr(NO3)2 is added to a mixture of F- and SO42-, we compare Qsp (the ion product) with Ksp. The salt with Qsp nearest to its Ksp will precipitate first. We compare Qsp for each salt using the initial concentrations provided and determine that SrSO4 will precipitate before SrF2. Using the Ksp values and the initial anion concentrations, we find that Sr2+ concentration at the start of precipitation is approximately 3.8x10-5 M for SrSO4.

When the second precipitate begins to form, we can calculate the percentage of the anion from the first precipitate remaining in solution by comparing the remaining concentration of that anion to its initial concentration before precipitation began.

Answer 2

a) The solubility product constant for SrSO4 is given by:

Ksp = [Sr2+][SO42-]

Let's assume the molar solubility of SrSO4 in water is x mol/L. Then at equilibrium, the concentrations of Sr2+ and SO42- ions will also be x mol/L each. Substituting these values into the expression for Ksp gives:

Ksp = x * x = x^2

So, x = sqrt(Ksp) = sqrt(7.610^-7) = 8.710^-4 mol/L

Therefore, the molar solubility of SrSO4 in pure water is 8.7*10^-4 mol/L.

b) The solubility product constant for SrF2 is given by:

Ksp = [Sr2+][F^-]^2

Let's assume the molar solubility of SrF2 in water is x mol/L. Then at equilibrium, the concentrations of Sr2+ and F- ions will also be x mol/L each. Substituting these values into the expression for Ksp gives:

Ksp = x * x^2 = x^3

So, x = (Ksp)^(1/3) = (7.910^-10)^(1/3) = 3.310^-4 mol/L

Therefore, the molar solubility of SrF2 in pure water is 3.3*10^-4 mol/L.

c) When Sr(NO3)2 is added to the solution containing F- and SO42-, two possible reactions can occur:

SrF2(s) ⇌ Sr2+(aq) + 2F-(aq)

SrSO4(s) ⇌ Sr2+(aq) + SO42-(aq)

We need to determine which salt will precipitate first. This can be done by calculating the ion product (Q) for each salt and comparing it to the corresponding solubility product constant (Ksp).

For SrF2: Q = [Sr2+][F^-]^2 = (0.020+ x)^2 * (2x)^2

For SrSO4: Q = [Sr2+][SO42-] = (0.10 + x) * x

where x is the molar solubility of the salt that will precipitate.

When the first salt starts to precipitate, Q = Ksp for that salt. Let's assume that SrF2 precipitates first. Then, we have:

Q = (0.020+ x)^2 * (2x)^2 = Ksp for SrF2 = 7.9*10^-10

Solving for x gives x = 2.2*10^-5 mol/L, which is the molar solubility of SrF2 at the point of precipitation.

The concentration of Sr2+ in the solution at this point is:

[Sr2+] = 0.020 + x = 0.020 + 2.2*10^-5 = 0.020022 mol/L

d) When the second salt starts to precipitate, the concentration of Sr2+ in the solution will remain the same, but the concentrations of F- and SO42- will change due to the reaction:

SrF2(s) + SrSO4(s) ⇌ 2Sr2+(aq) + SO42-(aq) + 2F-(aq)

Let's assume that SrF2 is the first precipitate and SrSO4 is the second. At the point when SrSO4 starts to precipitate, the concentration of F- in the solution is:

[F^-]

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