## General Chemistry: Principles and Modern Applications (10th Edition)

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.