Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 17 Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria - Study Questions - Page 677c: 68

Answer

Since $Q_{sp} \gt K_{sp}$, there will be precipitation of $PbCl_2$.

Work Step by Step

1. Calculate the molar mass $(NaCl)$: 22.99* 1 + 35.45* 1 = 58.44g/mol 2. Calculate the number of moles $(NaCl)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 1.20}{ 58.44}$ $n(moles) = 0.0205$ 3. Find the concentration in mol/L $(NaCl)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $ C(mol/L) = \frac{ 0.0205}{ 0.095} $ $C(mol/L) = 0.216$ 4. Write the $K_{sp}$ expression: $ PbCl_2(s) \lt -- \gt 1Pb^{2+}(aq) + 2Cl^{-}(aq)$ $ K_{sp} = [Pb^{2+}]^ 1[Cl^{-}]^ 2$ 5. Find the $Q_{sp}$ value $ Q_{sp} = (1.2 \times 10^{-3})^ 1 \times (0.216)^ 2$ $ Q_{sp} = (1.2 \times 10^{-3})^ 1 \times (0.216)^ 2$ $ Q_{sp} = 1.2 \times 10^{-3} \times (0.0467)$ $ Q_{sp} = 5.60 \times 10^{-5}$ $K_{sp} (PbCl_2) = 1.7 \times 10^{-5}$ Since $Q_{sp} \gt K_{sp}$, there will be precipitation
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.