Introductory Chemistry (5th Edition)

Published by Pearson
ISBN 10: 032191029X
ISBN 13: 978-0-32191-029-5

Chapter 15 - Chemical Equilibrium - Exercises - Cumulative Problems - Page 570: 99

Answer

Yes, there will be $CaSO_4$ precipitation.

Work Step by Step

1. Calculate the molar mass $(Na_2SO_4)$: 22.99* 2 + 32.07* 1 + 16* 4 ) = 142.05g/mol 2. Calculate the number of moles $(Na_2SO_4)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 0.105}{ 142.05}$ $n(moles) = 7.392\times 10^{- 4}$ 3. Find the concentration in mol/L $(Na_2SO_4)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $ C(mol/L) = \frac{ 7.392\times 10^{- 4}}{ 0.1000} $ $C(mol/L) = 7.392\times 10^{- 3}$ Therefore, the $[SO_4^{2-}]$ is equal to $7.392 \times 10^{-3}M$ 4. Calculate the $Q_{sp}$: - $K_{sp}$ expression: $[Ca^{2+}][SO_4^{2-}]$ $Q_{sp} = 0.025 \times 7.392 \times 10^{-3} = 1.8 \times 10^{-4}$ - The $K_{sp}$ for this salt is equal to $7.10 \times 10^{-5}$ Therefore, since the calculated value is greater, 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.