Answer
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:
$H_2O, CH_3COO^-, Na^+, CH_3COOH, H_3O^+, OH^-$
