Answer
(a) $ K_{sp} (SrF_2) = (7.8 \times 10^{-10})$
(b) $ K_{sp} (Ag_3PO_4)= (1.8 \times 10^{-18})$
Work Step by Step
(a)
1. Calculate the molar mass $(SrF2)$:
87.62* 1 + 19* 2 = 125.62g/mol
2. Calculate the number of moles $(SrF2)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{0.073}{ 125.62}$
$n(moles) = 5.8\times 10^{- 4}$
3. Find the concentration in mol/L $(SrF2)$:
$5.8 \times 10^{-4}$ mol in 1L: $5.8 \times 10^{-4} M (SrF2)$
4. Write the $K_{sp}$ expression:
$ SrF_2(s) \lt -- \gt 1Sr^{2+}(aq) + 2F^-(aq)$
$ K_{sp} = [Sr^{2+}]^ 1[F^-]^ 2$
5. Determine the ion concentrations:
$[Sr^{2+}] = [SrF_2] * 1 = [5.8 \times 10^{-4}] * 1 = 5.8 \times 10^{-4}$
$[F^-] = [SrF_2] * 2 = 1.2 \times 10^{-3}$
6. Calculate the $K_{sp}$:
$ K_{sp} = (5.8 \times 10^{-4})^ 1 \times (1.2 \times 10^{-3})^ 2$
$ K_{sp} = (5.8 \times 10^{-4}) \times (1.4 \times 10^{-6})$
$ K_{sp} = (7.8 \times 10^{-10})$
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(b)
1. Calculate the molar mass $(Ag3PO4)$:
107.87* 3 + 30.97* 1 + 16* 4 ) = 418.58g/mol
2. Calculate the number of moles $(Ag3PO4)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{6.7 \times 10^{-3}}{ 418.58}$
$n(moles) = 1.6\times 10^{- 5}$
3. Find the concentration in mol/L $(Ag3PO4)$:
$1.6 \times 10^{-5}$ mol in 1L: $1.6 \times 10^{-5} M (Ag3PO4)$
4. Write the $K_{sp}$ expression:
$ Ag_3PO_4(s) \lt -- \gt 3Ag^{+}(aq) + 1P{O_4}^{3-}(aq)$
$ K_{sp} = [Ag^{+}]^ 3[P{O_4}^{3-}]^ 1$
5. Determine the ion concentrations:
$[Ag^{+}] = [Ag_3PO_4] * 3 = [1.6 \times 10^{-5}] * 3 = 4.8 \times 10^{-5}$
$[P{O_4}^{3-}] = [Ag_3PO_4] * 1 = 1.6 \times 10^{-5}$
6. Calculate the $K_{sp}$:
$ K_{sp} = (4.8 \times 10^{-5})^ 3 \times (1.6 \times 10^{-5})^ 1$
$ K_{sp} = (1.1 \times 10^{-13}) \times (1.6 \times 10^{-5})$
$ K_{sp} = (1.8 \times 10^{-18})$
