Answer
(a) $pH (Buffer) = 4.54$
(b) The pH of the buffer is higher. (Because we added a base.)
$pH (Lactic acid)= 2.43$
Work Step by Step
1. Calculate the molar mass:
22.99* 1 + 12.01* 1 + 1.01* 3 + 12.01* 1 + 1.01* 1 + 16* 1 + 1.01* 1 + 12.01* 1 + 16* 2 = 112.07g/mol
2. Calculate the number of moles
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{2.75}{ 112.07}$
$n(moles) = 0.0245$
3. Find the concentration in mol/L:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 0.0245}{ 0.05} $
$C(mol/L) = 0.491$
4. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 1.4 \times 10^{- 4})$
$pKa = 3.85$
5. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{0.491}{0.1}$
- 4.91: It is.
6. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{0.491}{1.4 \times 10^{-4}} = 3505$
- $ \frac{0.1}{1.4 \times 10^{-4}} = 714$
7. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 3.85 + log(\frac{0.491}{0.1})$
$pH = 3.85 + log(4.91)$
$pH = 3.85 + 0.691$
$pH = 4.54$
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Calculating the pH of the pure acid solution:
1. We have those concentrations at equilibrium:
-$[H_3O^+] = [CH_3CHOHC{O_2}^-] = x$
-$[CH_3CHOHCO_2H] = [CH_3CHOHCO_2H]_{initial} - x = 0.1 - x$
For approximation, we consider: $[CH_3CHOHCO_2H] = 0.1M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][CH_3CHOHC{O_2}^-]}{ [CH_3CHOHCO_2H]}$
$Ka = 1.4 \times 10^{- 4}= \frac{x * x}{ 0.1}$
$Ka = 1.4 \times 10^{- 4}= \frac{x^2}{ 0.1}$
$ 1.4 \times 10^{- 5} = x^2$
$x = 3.74 \times 10^{- 3}$
Percent dissociation: $\frac{ 3.74 \times 10^{- 3}}{ 0.1} \times 100\% = 3.74\%$
%dissociation < 5% : Right approximation.
Therefore: $[H_3O^+] = [CH_3CHOHC{O_2}^-] = x = 3.74 \times 10^{- 3}M $
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 3.74 \times 10^{- 3})$
$pH = 2.43$
