What are [Ba2+] and [CrO42−] in a saturated BaCrO4 solution if the Ksp of BaCrO4 is 1×10−10?

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Question:

What are [Ba2+] and [CrO42−] in a saturated BaCrO4 solution if the Ksp of BaCrO4 is 1×10−10?

[Ba2+] =
[CrO42−] =

Answer:

1)  Chemical equation of the equilibrium

BaCrO4⇄ Ba (2+) + CrO4 (2-)

2) Ksp equation

Ksp = [Ba(2+)] [CrO4 (2-)]

where Ksp = 1.0 * 10^ – 10 and [Ba(2+)] = [CrO4(2-)] = x

3) Calculations

1.0 * 10^ -10 = x * x

=> x^2 = 1.10 * 10^-10

=> x = √[1.0 * 10^ -10] = 0.000010

Answer: [Ba(2+)] = [CrO4(2-)] = 0.00001 M

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