Question

A 24.00 mL sample of a solution of Pb(ClO3)2 was diluted with water to 52.00 mL. A 17.00 mL sample of the dilute solution was found to contain 0.220 M ClO3−(aq). What was the concentration of Pb(ClO3)2 in the original undiluted solution? 3.60 × 10−2 M 7.19 × 10−2 M 0.238 M 0.156 M 0.477 M

Answer 1

0.238 M

Step-by-step explanation:

A 17.00 mL sample of the dilute solution was found to contain 0.220 M ClO₃⁻(aq). The concentration is an intensive property, so the concentration in the 52.00 mL is also 0.220 M ClO₃⁻(aq). We can find the initial concentration of ClO₃⁻ using the dilution rule.

C₁.V₁ = C₂.V₂

C₁ × 24.00 mL = 0.220 M × 52.00 mL

C₁ = 0.477 M

The concentration of Pb(ClO₃)₂ is:

`(0.477molClO_(3)^(-) )/(L) * (1molPb(ClO_(3))_(2))/(2molClO_(3)^(-)) =0.238M`