Answer
No, the compound does not completely dissolve. Only 0.012 g (12 mg) of it does.
Work Step by Step
1. Calculate the molar mass $(PbSO_4)$:
207.2* 1 + 32.07* 1 + 16* 4 ) = 303.27g/mol
2. Calculate the number of moles $(PbSO_4)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 0.055}{ 303.27}$
$n(moles) = 1.8\times 10^{- 4}$
3. Find the concentration in mol/L $(PbSO_4)$:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 1.8\times 10^{- 4}}{ 0.25} $
$C(mol/L) = 7.3\times 10^{- 4}$
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Now, find the solubility:
4. Write the $K_{sp}$ expression:
$ PbSO_4(s) \lt -- \gt 1Pb^{2+}(aq) + 1S{O_4}^{2-}(aq)$
$2.5 \times 10^{-8} = [Pb^{2+}]^ 1[S{O_4}^{2-}]^ 1$
5. Considering a pure solution: $[Pb^{2+}] = 1S$ and $[S{O_4}^{2-}] = 1S$
$2.5 \times 10^{-8}= ( 1S)^ 1 \times ( 1S)^ 1$
$2.5 \times 10^{-8} = 1S^ 2$
$2.5 \times 10^{-8} = S^ 2$
$ \sqrt [ 2] {2.5 \times 10^{-8}} = S$
$1.6 \times 10^{-4} = S$
- This is the molar solubility value for this salt.
** Since the concentration added is greater than the solubility value, there will be precipitation, and only $1.6 \times 10^{-4}M$ will dissolve.
6. Find the number of moles:
$Concentration(M) = \frac{n(mol)}{V(L)}$
$1.6 \times 10^{-4} = \frac{n(mol)}{0.25}$
$1.6 \times 10^{-4} * 0.25 = n(mol)$
$4.0 \times 10^{-5} moles = n(mol)$
8. Calculate the mass
$mm(g/mol) = \frac{mass(g)}{n(mol)}$
$mm(g/mol) * n(mol) = mass(g)$
$ 303.27 * 4 \times 10^{-5} = mass(g)$
$0.012 = mass(g)$
