Answer
The solubility is about 2.057g/L.
Work Step by Step
1. Calculate $[H^+]$
$[H^+] = 10^{-pH}$
$[H^+] = 10^{- 4.53}$
$[H^+] = 2.95 \times 10^{- 5}$
2. Calculate the Ka
$Ka = 10^{-pKa} = 10^{-7.23} = 5.89 \times 10^{-8}$
3. Use the Ka equation to find the concentration of the acid:
$5.89 \times 10^{-8} = \frac{(2.95 \times 10^{-5})^2}{[Acid]}$
$[Acid] = \frac{8.70 \times 10^{-10}}{5.89 \times 10^{-8}} = 1.48 \times 10^{-2}M$
4. Convert the number of moles in grams:
- Molar mass: 12*6 (C) + 1*5(H) + 14*1 (N) + 16*3(O) = 139g /mol
mm (g/mol) = $\frac{mass(g)}{n(moles)}$
139 = $\frac{mass(g)}{ 0.0148}$
mass(g) = 139$\times$ 0.0148
mass(g) = 2.0572g
Therefore: 0.0148 mol/L = 2.057g/L
For satured solutions, the solubility of the compound is its concentration.
