Chemistry Solubility Math Question 9

Question

How many mg of CuNO₃ can be dissolved in 124.0 mL of 1E-2 M RbBr before a precipitate forms? K_sp(CuBr)=4.2x10^-8

Answer 1

CuBr(s) < ==> Cu⁺(aq) + Br^-(aq)

Ksp = 4.2x10^-8 = [Cu⁺][Br-]

4.2x10^-8 = (x)(1x10^-2)

x = [Cu⁺] = 4.2x10^-6 M

To find grams (mg) of CuNO₃ in 124.0 ml that produces 4.2x10^-6 M, we proceed as follows:

molar mass CuNO₃ = 188 g/mole

124.0 ml = 0.124 L

x mole/0.124 L = 4.2x10^-6 mol/L

x = 5.21x10^-7 moles CuNO₃

5.21x10^-7 moles CuNO₃ x 188 g/mole = 9.79x10^-5 g

9.79x10^-5 g CuNO₃ x 1000 mg/g = 9.79x10^-2 mg = 0.0979 mg CuNO₃