Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 15 - Additional Aqueous Equilibria - Questions for Review and Thought - General Questions - Page 693d: 85a


Molecules and ions in order of decreasing concentration: $H_2O, CH_3COO^-, Na^+, CH_3COOH, H_3O^+, OH^-$.

Work Step by Step

- This is an aqueous solution, so, there are $H_2O$, $H_3O^+$ and $OH^-$ molecules/ions. - $NaCH_3COO$ is a salt, so, there are $Na^+$ and $CH_3COO^-$ in the solution. - There are $CH_3COOH$ molecules in the solution. ------ Now, we have to put them in decreasing concentration order. Before that, we need to know the salt concentration. 1. Calculate the molar mass $(NaCH_3COO)$: 22.99* 1 + 12.01* 1 + 1.01* 3 + 12.01* 1 + 16* 1 + 16* 1 = 82.04g/mol 2. Calculate the number of moles $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 4.95}{ 82.04}$ $n(moles) = 0.06034$ 3. Find the concentration in mol/L: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $ C(mol/L) = \frac{ 0.06034}{ 0.25} $ $C(mol/L) = 0.241$ - The highest concentration is $H_2O$, because it is the solvent. - $[CH_3COOH] = 0.150M$, and $[NaCH_3COO] = 0.241M$, so: - $CH_3COO^-*** \gt Na^+$ $ \gt CH_3COOH$ *** Notice, there are some $CH_3COO^-$ being produced by the dissociation of $CH_3COOH$. And now we have $H_3O^+$ and $OH^-$: Since $K_a (CH_3COOH) \gt K_b(CH_3COO^-)$, and they have close concentrations values, the solution is going to be acidic: $[H_3O^+] > [OH^-]$ Final order: $H_2O, CH_3COO^-, Na^+, CH_3COOH, H_3O^+, OH^-$
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