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## Homework Statement

I have worked these two problems to the best of my knowledge and I keep getting the wrong answer (I'm not sure what the right answer is). Here are the questions:

1. At a certain temperature the solubility of barium chromate (BaCrO4) is 1.9×10-5 mol/L. What is the Ksp of BaCrO4 at this temperature?

2. What is the molar solubility of silver phosphate (Ag3PO4) in water? Ksp = 1.8×10-18.

## Homework Equations

The relevant equations are Ksp=([products]/[reactants])

## The Attempt at a Solution

For question 1: BaCrO

_{4}⇔Ba

^{2+}+ CrO

_{4}

^{2-}

K

_{sp}=[Ba

^{2+}][CrO

_{4}

^{2-}]/1 (since it is a solid)

K

_{sp}=[x][x]=x

^{2}

K

_{sp}=[1.9x10

^{-5}]

^{2}=3.61x10

^{-10}= 3.6x10

^{-10}

For question 2: Ag

_{3}PO

_{4}⇔3Ag

^{+}+ PO

_{4}

^{3-}

1.8x10

^{-18}= ([Ag

^{+}]

^{3}[PO

_{4}

^{3-}])/1

1.8x10

^{-18}= [3x]

^{3}[x]

1.8x10

^{-18}= [27x

^{4}] then divide both sides by 27

6.6666667x10

^{-20}= x

^{4}then do the fourth root on both sides

x = 1.6068568x10

^{-5}= 1.6x10

^{-5}

I would very much appreciate any insight into this. Thank you!