Answer
It is necessary for about 13.4g of the acid to form this buffer.
Work Step by Step
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
$mass(g) = mm(g/mol) * n(moles)$
$mass(g) = 122.13 * 0.1098$
$mass(g) = 13.41$
