100 mL of 0.0500 M lead(II) nitrate and 200.0 mL of 0.100 M sodium iodide are mixed. Calculate the final concentrations of all ions and the mass of any precipitate. Assume K_sp for PbI₂ = 1.4 x 10^-8.

Need work/steps/concepts, pls make sure answer matches. Ans: [Pb⁺²] = 1.3 x 10^-5M , [I^-] = 3.33 x 10^-2M, [Na⁺] = 0.667 M; [NO₃^-] = 0.0333M, mass PbI₂ = 2.31 g

Question

100 mL of 0.0500 M lead(II) nitrate and 200.0 mL of 0.100 M sodium iodide are mixed. Calculate the final concentrations of all ions and the mass of any precipitate. Assume K_sp for PbI₂ = 1.4 x 10^-8.

Need work/steps/concepts, pls make sure answer matches. Ans: [Pb⁺²] = 1.3 x 10^-5M , [I^-] = 3.33 x 10^-2M, [Na⁺] = 0.667 M; [NO₃^-] = 0.0333M, mass PbI₂ = 2.31 g

Unlock all answers

Related textbook solutions

Basic Chemistry

Principles of Chemistry Molecular Approach

Principles of Chemistry Molecular Approach

Chemistry: Structure and Properties

Chemistry: A Molecular Approach

Chemistry: A Molecular Approach

Principles of Chemistry: A Molecular Approach

Chemistry: The Central Science

Weekly leaderboard