Answer
(a) $ K_{sp} (CdC{O_3}) = (1.0 \times 10^{-12})$
(b) $ K_{sp} (Ca(OH)_2) = (4.76 \times 10^{-6})$
(c) $ K_{sp} (PbBr_2)= (6.62 \times 10^{-6})$
(d) $ K_{sp} (BaCr{O_4})= (1.1 \times 10^{-10})$
Work Step by Step
(a)
1. Write the $K_{sp}$ expression:
$ CdCO_3(s) \lt -- \gt 1Cd^{2+}(aq) + 1{CO_3}^{2-}(aq)$
$ K_{sp} = [Cd^{2+}]^ 1[{CO_3}^{2-}]^ 1$
2. Determine the ion concentrations:
$[Cd^{2+}] = [CdCO_3] * 1 = [1 \times 10^{-6}] * 1 = 1 \times 10^{-6}$
$[{CO_3}^{2-}] = [CdCO_3] * 1 = 1 \times 10^{-6}$
3. Calculate the $K_{sp}$:
$ K_{sp} = (1 \times 10^{-6})^ 1 \times (1 \times 10^{-6})^ 1$
$ K_{sp} = (1 \times 10^{-6}) \times (1 \times 10^{-6})$
$ K_{sp} = (1.0 \times 10^{-12})$
(b)
1. Write the $K_{sp}$ expression:
$ Ca(OH)_2(s) \lt -- \gt 1Ca^{2+}(aq) + 2{OH}^{-}(aq)$
$ K_{sp} = [Ca^{2+}]^ 1[{OH}^{-}]^ 2$
2. Determine the ions concentrations:
$[Ca^{2+}] = [Ca(OH)_2] * 1 = [0.0106] * 1 = 0.0106$
$[{OH}^{-}] = [Ca(OH)_2] * 2 = 0.0212$
3. Calculate the $K_{sp}$:
$ K_{sp} = (0.0106)^ 1 \times (0.0212)^ 2$
$ K_{sp} = (0.0106) \times (4.49 \times 10^{-4})$
$ K_{sp} = (4.76 \times 10^{-6})$
(c)
1. Calculate the molar mass:
207.2* 1 + 79.9* 2 = 367g/mol
2. Calculate the number of moles
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{4.34}{ 367}$
$n(moles) = 0.0118$
3. Find the concentration in mol/L:
$0.0117$ mol in 1L: $0.0117 M$
4. Write the $K_{sp}$ expression:
$ PbBr_2(s) \lt -- \gt 1Pb^{2+}(aq) + 2{Br}^{-}(aq)$
$ K_{sp} = [Pb^{2+}]^ 1[{Br}^{-}]^ 2$
5. Determine the ion concentrations:
$[Pb^{2+}] = [PbBr_2] * 1 = [0.0117] * 1 = 0.0117$
$[{Br}^{-}] = [PbBr_2] * 2 = 0.0237$
6. Calculate the $K_{sp}$:
$ K_{sp} = (0.0117)^ 1 \times (0.0237)^ 2$
$ K_{sp} = (0.0117) \times (5.59 \times 10^{-4})$
$ K_{sp} = (6.62 \times 10^{-6})$
(d)
1. Calculate the molar mass:
137.3* 1 + 52* 1 + 16* 1 + 16* 4 ) = 269.3g/mol
2. Calculate the number of moles
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{2.8 \times 10^{-3}}{ 269.3}$
$n(moles) = 1\times 10^{- 5}$
3. Find the concentration in mol/L:
$1 \times 10^{-5}$ mol in 1L: $1 \times 10^{-5} M$
4. Write the $K_{sp}$ expression:
$ BaCr{O_4}(s) \lt -- \gt 1Ba^{2+}(aq) + 1{Cr{O_4}}^{2-}(aq)$
$ K_{sp} = [Ba^{2+}]^ 1[{Cr{O_4}}^{2-}]^ 1$
5. Determine the ionconcentrations:
$[Ba^{2+}] = [BaCr{O_4}] * 1 = [1 \times 10^{-5}] * 1 = 1 \times 10^{-5}$
$[{Cr{O_4}}^{2-}] = [BaCr{O_4}] * 1 = 1 \times 10^{-5}$
6. Calculate the $K_{sp}$:
$ K_{sp} = (1 \times 10^{-5})^ 1 \times (1 \times 10^{-5})^ 1$
$ K_{sp} = (1 \times 10^{-5}) \times (1 \times 10^{-5})$
$ K_{sp} = (1.1 \times 10^{-10})$
