Use the molar solubility 1.08×10^-5 in pure water to calculate Ksp for BaCrO4.?

a.) Use the molar solubility 1.08×10^-5 in pure water to calculate Ksp for BaCrO4.

Answer 1

a) BaCrO4 –> Ba2+ + CrO42-. The expression for Ksp is [Ba2+][CrO42-], and since we have 1.08 x 10^-5 M BaCrO4, each of those concentrations is 1.08 x 10^-5. Hence Ksp is (1.08 x 10^-5)^2 = 1.17 x 10^-10

b) Same procedure, except Ag2SO3 –> 2Ag+ + SO3(2-), so the Ksp expression is [Ag+]^2[SO3(2-)].

c) Same procedure, except Pd(SCN)2 –> Pd(2+) + 2SCN-. You should be able to follow the procedure by now to guess what the Ksp expression is and what numbers to put in to compute its value.

Answer 2

a. 1 BaCrO4 ionizes to producw one Ba ion and 1 CrO4 ion. If BaCrO4 solubility is 1.08 X 10^-5, then that means that the concentration of each ion is also 1.08 x 10^-5

Ksp = [Ba++][CRO4=] = (1.08 X 10^-5)(1.08 X 10^-5)

Ksp = 1.1664 x 10^-10

the others are similar

Answer 3

BaCrO4 –> Ba2+ + CrO4 2-, NO COEFFICIENTS, the 4 of CrO4 tells you that there are 4 O in CrO4 2- ion, NOT 4 CrO ions Ksp = [Ba2+][CrO4 2-] molar solubility = 1.08x^-5M = [Ba2+] Ksp = 1.08×10^-5^2 = 1.17×10^-10

Answer 4

Bacro4

Answer 5

The answers for all of them are
a)1.17*10^-10
b)1.49*10^-14
c)4.38*10^-23

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