General Chemistry I1 Test S Take Home Aqueous Equilibria Part 3 Solubility Instr
ID: 571838 • Letter: G
Question
General Chemistry I1 Test S Take Home Aqueous Equilibria Part 3 Solubility Instructions: Place the letter corresponding to the correct answer on the line next to the question number then in the space below the problem provide a logical reason or simple calculation defending your answer Section 1 Multiple Choice (4 pts each- 1 pt for answer 3 for the reason) Which of the following compounds will have the highest molar solubility in pure water? A) POS04. Ksp-1.82 × 10-8 1 pt 1. C) Agl, Ksp-8.51 10-17 Mtgcos, Ksp-6.82 10-6DPb, Ksp 9.0410-29 3pts Reason: Which of the following compounds will be more soluble in acidic solution than in pure water? A) PbCl2 B) Fes 1 pt - C) Ca(CI04)2 D) Cul 3 ptsReason: 1 pt 3. In which of the following solutions would solid PbCl2 be expected to be the least soluble at 25 C? A) 0.1 M HCI B) 0.1 M NaNO, C) 0.1 M CaCl2 D) 0.1 M Pb(NO3)2 3 pts. Reason: Section 2 Show your work 5 pts 1. Determine the molar solubility of BaF2 in pure water Ksp for BaF2 = 2.45 × 10-5. 3 pts 2The molar solubility of Cul is 2.26 x 10-6 M in pure water. Calculate the Ksp for Cul Paoe 7 of 7Explanation / Answer
1. Higher the Ksp value higher is the solubility. Since Ksp of MgCO3 is highest its solubility will be highest too. Hence correct choice is C
2. The dalt that has anion coming from weak acid will combine with protons released by string acid and hence will be more soluble. In this case choice b is correct because (ClO4)2 is the only one coming from weak acid.
3. PbCl2 will have least solubility in that solution which has common ion with PbCl2 . So choice b is correct,
4.
BaF2 (aq)-> Ba2+ + 2 F-
let sulubility = x
Ksp =x* (2x)2 /[BaF2]
BaF2 = 1 because activity of BaF2 (aq) = 1
x= 0.0182 M
5.
CuI-> Cu++ F-
solubility =x= 2.26*10-6 M
Ksp = x2
Ksp = 5.1*10-12
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