Question

The solubility product ($K_{sp}$) of Ca(OH)₂ at 25ºC is 4.42 × 10⁻⁵. A 500 ml. of saturated solution of Ca(OH)₂ is mixed with equal volume of 0.4 M NaOH. How much Ca(OH)₂ in milligrams is precipitated?

Answer 1

Answer:

Let S M be the solubility of Ca(OH)

2

in pure water.

S

Ca(OH)

2

⇌

S

Ca

2+

2S

OH

−

K

sp

=[Ca

2+

][OH

−

]

Substitute values in the above expression.

4.42×10

−5

=S(2S)

2

or S=0.0223M

500 ml of solution contains 0.01115 moles of calcium ions.

The molarity of NaOH solution is 0.4 M. When it is mixed with equal volume of calcium hydroxide solution, its molarity is reduced to half, i.e., 0.2M.

[OH

−

]=0.2M

[Ca

2+

]=

[OH

−

]

2

K

sp

=

(0.2)

2

4.42×10

−5

=0.001105M.

The number of moles of calcium ion precipitated =0.01115−0.001105=0.010045.

The number of milimoles of calcium hydroxide precipitated =0.010045×74×1000=743 mg.