Answered

At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Get detailed and accurate answers to your questions from a community of experts on our comprehensive Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

250 mL of a solution of calcium oxalate is the evaporated until only a residue of solid calcium
oxalate is left. The mass of the residue recovered from the container was 154 mg. What is the
solubility product of calcium oxalate?


Sagot :

Answer:

2.3 * 10^-5

Explanation:

Recall that the solubility of a solute is the amount of solute that dissolves in 1 dm^3 or 1000cm^3 of solution.

Hence;

Amount of calcium oxalate = 154 * 10^-3/128.097 g/mol = 1.2 * 10^-3 mols

From the question;

1.2 * 10^-3 mols dissolves in 250 mL

x moles dissolves in 1000mL

x = 1.2 * 10^-3 mols * 1000/250

x= 4.8 * 10^-3 moldm^-3

CaC2O4(s) ------->Ca^2+(aq) + C2O4^2-(aq)

Hence Ksp = [Ca^2+] [C2O4^2-]

Where;

[Ca^2+] = [C2O4^2-] = 4.8 * 10^-3 moldm^-3

Ksp = (4.8 * 10^-3)^2

Ksp = 2.3 * 10^-5

Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.