Question

what is the maximum mass of calcium nitrate that can be added to 100.0mL of 1.0M ammonia (kb=1.8x10^-5) without precipitating calcium hydroxide (ksp=5.5x10^-6)? assume no volume changes occurred upon the addition.

Answer 1

The solutionis ammonia solution of concentration = 1.0M

thus pOH of the solution = 1/2 [pKb -logc]

= 1/2 [4.75 -log1]

= 2.375

Thus [OH-] in solution = antilog (-2.375)

=4.127x10^-3 M

Now if calcium nitrate is added the precipitation of Ca(OH)2 should not occur

the solubility equilibrium is

Ca(OH)2(s) <---------> Ca+2 (aq) + 2OH- (aq)

- s   4.127x10^-3 M

Thus Ksp = s (4.127x10^-3 M)² = 5.5 x10^-6

thus s = 5.5 x10^-6  /(4.127x10^-3 )²

   ⁼0.322M

thus the maxiumum molarity of solution in Ca+2 can be 0.322M before Ca(OH)2 starts precipitation.

0.322 moles of Ca(NO3)2 in 1L of solution

? in 100mL of solution

= 0.0322 moles in 100mL

thus mass of calcium nitrate that can be added before precipitating calcium hydroxide =

= moles x molar mass

= 0.0322x 164g/mol

= 5.2808 g