What is the molar concentration of Pb+ in a solution that contains 6.73 ppm of Pb(NO3)2?
Answer with correct number of S. F.
(Hint: the unit of your answer is mol/L = M)

2.03 × 10⁻⁵ M

Explanation:

Step 1: Given data

Concentration of Pb(NO₃)₂: 6.73 ppm = 6.73 mg/L

Step 2: Convert 6.73 mg/L to mol/L

The molar mass of 331.2 g/mol.

6.73 × 10⁻³ g/L × 1 mol/331.2 g = 2.03 × 10⁻⁵ mol/L = 2.03 × 10⁻⁵ M

Step 3: Calculate the molar concentration of Pb²⁺

Let's consider the ionization of Pb(NO₃)₂.

Pb(NO₃)₂(aq) ⇒ Pb²⁺(aq) + 2 NO₃⁻(aq)

The molar ratio of Pb(NO₃)₂ to Pb²⁺ is 1:1. The molar concentration of Pb²⁺ is 1/1 × 2.03 × 10⁻⁵ M = 2.03 × 10⁻⁵ M.

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