Question

What is the concentration of barium ions available in solution if you dissolve 150g of barium...

What is the concentration of barium ions available in solution if you dissolve 150g of barium sulfate. (Ksp = 1.1 x 10^-10) in 750 mL of water?

Homework Answers

Answer #1

Lets calculate the concentration of BaSO4

Molar mass of BaSO4 = 1*MM(Ba) + 1*MM(S) + 4*MM(O)

= 1*137.3 + 1*32.07 + 4*16.0

= 233.37 g/mol

mass of BaSO4 = 150 g

we have below equation to be used:

number of mol of BaSO4,

n = mass of BaSO4/molar mass of BaSO4

=(150.0 g)/(233.37 g/mol)

= 0.6428 mol

volume , V = 750 mL

= 0.75 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 0.6428/0.75

= 0.857 M

The salt dissolves as:

BaSO4 <----> Ba2+ + SO42-

   s s

Ksp = [Ba2+][SO42-]

1.1*10^-10=(s)*(s)

1.1*10^-10= 1(s)^2

s = 1.049*10^-5 M

Since solubility is less than concentration,

1.049*10^-5 M will be part of solution

Answer: 1.05*10^-5 M

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