ntent/0/1102-HW4.pdf 3) What is the Kip for the following compounds if the given

Question

ntent/0/1102-HW4.pdf 3) What is the Kip for the following compounds if the given masses can dissolve in 100. mlL of water a) Strontium fluoride (1.2 x 10 g/100. ml) b) Lithium acetate (40.8 g/100. ml) 4) Lithium fluoride has a Kup of 1.8 x 10. If 2.50 g of lithium fluoride is placed in a beaker containing 500. mL of water, what are the concentrations of the ions in solution after the solution is at equilibrium? 5) 50.0 ml of 1.00 M sodium acetate and 50.0 mL of 1.00 M silver() nitrate are mixed together. What mass of silverÜ) acetate precipitates from solution? Kip for silver() acetate 3.7x10

Explanation / Answer

Ans 3

Part a

Solubility of strontium flouride = 1.20*10^-4g / 100 mL

= 0.0000012 g/mL

Molar solubility = (0.0000012 g/mL) / (125.62 g/mol)

= 9.552 x 10^-9 mol/mL x 1000mL/L

s = 9.552 x 10^-6 mol/L

SrF2(s) <---> Sr2+(aq) + 2 F-(aq)

Equilibrium constant expression of the reaction

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

Ksp = s(2s)^2 = 4s^3

Ksp = 4*(9.552 x 10^-6)^3
= 3.486 x 10^-15

Part b

Solubility of Lithium acetate = 40.8g / 100 mL

= 0.408 g/mL

Molar solubility = (0.408 g/mL) / (65.984 g/mol)

= 0.00618 mol/mL x 1000mL/L

s = 6.18 mol/L

CH3COOLi = Li+ + CH3COO-

Equilibrium constant expression of the reaction

Ksp = [Li+][CH3COO-]

Ksp = s(s) = s^2

Ksp = (6.18)^2
= 38.23

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