Calculate the solubility at 25°c of cubr in pure water and in a 0.0030m cobr2 solution. you'll find ksp data in the aleks data tab. round both of your answers to 2 significant digits. solubility in pure water: gl

the solubility at 25°c of cubr in pure water = 0.011 g/L

The solubility at 25°c of cubr  in a 0.0030m cobr2 solution = 0.00014 g/L

Explanation:

Step 1: Data given

Ksp of CuBr = 6.27 × 10^-9

Molar mass CuBr = 143.45 g/mol

Step 2: Calculate the solubility at 25°c of cubr in pure water

Ksp = [Cu+][Br-]

The initial concentration

[Cu+] = 0M

[Br-] = 0M

The concentration at the equilibrium

[Cu+] = X M

[Br-] = X M

Ksp =6.27 * 10^-9 =  X * X = X²

S = 7.9 *10^-5 mol /l  

7.9 * 10^-5 mol/L * 143.45 g/mol = 0.011 g/L

Step 3: Calculate the solubility at 25°c of cubr  in a 0.0030m cobr2 solution.

The balanced equation:

CoBr2(aq) → Co^2+(aq) + 2 Br⁻(aq)

For 1 mol CoBr2 we'll have 1 mol Co^2+ and 2 moles Br-

For 0.0030 M CoBr2 we'll have 0.0030 M Co^2+ and 0.0060 M Br-

The initial concentration

[Cu+] = 0M

[Br-] = 0.0060 M

At the equilibrium

[Cu+] = X

[Br-] = 0.0060 + X

Ksp = 6.27 * 10^-9 = X * (0.0060+X)

6.27 *10^-9 = 0.0060 X + X²

X = 1.0 * 10^-6 mol/L

1.0 * 10^-6 mol/L * 143.45 g/mol = 0.00014 g/L

i do not get it   at all