## Chemistry: The Molecular Science (5th Edition)

1. to 3.: $NaC_6H_5COO:$ 1. Calculate the molar mass: 22.99* 1 + 12.01* 6 + 1.01* 5 + 12.01* 1 + 16* 1 + 16* 1 = 144.11g/mol 2. Calculate the number of moles $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 14.4}{ 144.11}$ $n(moles) = 0.09992$ 3. Find the concentration in mol/L: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 0.09992}{ 1}$ $C(mol/L) = 0.09992$ 4. Calculate $[H_3O^+]$: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 3.88}$ $[H_3O^+] = 1.318 \times 10^{- 4}M$ 5. Write the $K_a$ equation, and find the acid concentration: $K_a = \frac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}$ $1.2 \times 10^{-4} = \frac{1.318 \times 10^{-4}*[0.09991]}{[C_6H_5COOH]}$ $1.2 \times 10^{-4} = \frac{1.317 \times 10^{-5}}{[C_6H_5COOH]}$ $C_6H_5COOH = \frac{1.317 \times 10^{-5}}{[1.2 \times 10^{-4}]}$ $1.317 \times 10^{-5}M = [C_6H_5COOH]$ $C_6H_5COOH:$ 6. Calculate the number of moles: $n(moles) = concentration(M) * volume(L)$ $n(moles) = 0.1098 * 1$ $n(moles) = 0.1098$ 7. Find the mass value in grams: 12.01* 6 + 1.01* 5 + 12.01* 1 + 16* 1 + 16* 1 + 1.01* 1 ) = 122.13g/mol $mass(g) = mm(g/mol) * n(moles)$ $mass(g) = 122.13 * 0.1098$ $mass(g) = 13.41$