Chemistry and Chemical Reactivity (9th Edition)

1. Calculate $[H_3O^+]$: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 4.5}$ $[H_3O^+] = 3.16 \times 10^{- 5}M$ 2. Write the $K_a$ equation, and find the ratio: $K_a = \frac{[H_3O^+][CH_3C{O_2}^-]}{[CH_3C{O_2}H]}$ $1.8 \times 10^{-5} = \frac{3.16 \times 10^{-5}*[CH_3C{O_2}^-]}{[CH_3C{O_2}H]}$ $\frac{1.8 \times 10^{-5}}{3.16 \times 10^{-5}} = \frac{[CH_3C{O_2}^-]}{[CH_3C{O_2}H]}$ $0.57 = \frac{[CH_3C{O_2}^-]}{[CH_3C{O_2}H]}$ 3. Calculate the needed concentration of the base: $0.57 = \frac{[CH_3C{O_2}^-]}{0.10}$ $0.57 * 0.10 = {[CH_3C{O_2}^-]}$ $0.057M = {[CH_3C{O_2}^-]}$ ** Notice: $[NaCH_3C{O_2}^-] = [CH_3C{O_2}^-] = 0.057M$ 4. Find the number of moles: $Concentration(M) = \frac{n(mol)}{V(L)}$ $0.057 = \frac{n(mol)}{1}$ $0.057 * 1 = n(mol)$ $0.057 moles = n(mol)$ 5. Determine the molar mass of this compound $(NaCH_3CO_2)$: ** Remember: we are adding sodium acetate, not just the base ion. 22.99* 1 + 12.01* 1 + 1.01* 3 + 12.01* 1 + 16* 2 = 82.04g/mol 6. Calculate the mass $mm(g/mol) = \frac{mass(g)}{n(mol)}$ $mm(g/mol) * n(mol) = mass(g)$ $82.04 * 0.057 = mass(g)$ $4.7 = mass(g)$