Question

What weight of lead nitrate in miligrams would needed to dissolve in 9990 ml water to prepare 75 ppm of pb2+?

Answer 1

x (Pb(NO₃)₂) = 75ppm = 0.0075% molar fraction

V(H₂O) = 9990 ml

m(Pb(NO₃)₂) - ? mg

x Pb(NO₃)₂ = $\frac{\text{n(Pb(NO3)2}}{\text{n(Pb(NO3)2) + n(H2O)}} \times 100\%$

n(H₂O) = V(H₂O)/Vₘ = 9.99L/22.4 L/mol = 0.45 mol

$\frac{\text{n(Pb(NO3)2}}{\text{n(Pb(NO3)2) + n(H2O)}} = \frac{0.0075\%}{100\%} = 7.5 \times 10^{-5}$

7.5 × 10⁻⁵ * n(Pb(NO₃)₂) + 7.5 × 10⁻⁵ * n(H₂O) = n(Pb(NO₃)₂)

(1 - 7.5 × 10⁻⁵) * n(Pb(NO₃)₂) = 7.5 × 10⁻⁵ * 0.45 mol

n(Pb(NO₃)₂) = 7.5 × 10⁻⁵ * 0.45 mol / (1 - 7.5 × 10⁻⁵) = 3.37 × 10⁻⁵ mol

m(Pb(NO₃)₂) = n(Pb(NO₃)₂) * M(Pb(NO₃)₂) = 3.37 × 10⁻⁵ mol * 331 g/mol = 1115 × 10⁻⁵ g = 11.15 mg