Q:

Preparation of acetate buffer from sodium acetate and hydrochloric acid

A $\pu{1.0 L}$ buffer solution of $\mathrm{pH}~4.43$ is made up of $\pu{0.30 M}$ sodium ethanoate and $\pu{0.20 M}$ $\ce{HCl}$ solutions. Calculate the volume of sodium ethanoate and $\pu{0.20 M}$ $\ce{HCl}$ solutions used to prepare this buffer solution.

The answer given for volume of $\ce{HCl}$ is $\pu{750 mL}$ and for the volume of volume $\ce{CH3COONa}$ is $\pu{250 mL}.$
My approach is to assume $\ce{HCl}$ as the limiting reactant and use ICE table to determine the equilibrium concentration of the ethanoic acid and sodium ethanoate at the end of the reaction. Then I use the Henderson–Hasselbalch equation.
The answer I got was $\pu{500 mL}$ for both the conjugate base and weak acid. Is the answer wrong?

A:

To answer this question, we have to assume that the given buffer solution is an ideal solution of which the total volume of it always equals to the summed volumes of sodium ethanoate ($\ce{NaOAc}$) and $\ce{HCl}$ solutions when they are added together. It should also be assumed that $\mathrm{p}K_\mathrm{a}$ of ethanoic acid (acetic acid; $\ce{HOAc}$) is given ($\mathrm{p}K_\mathrm{a} = 4.76$).
When added together, $\ce{HCl}$ (a strong acid) and $\ce{NaOAc}$ (weekly basic salt) react completely to give $\ce{HOAc}$ and $\ce{NaCl}$, according to the following equation:
$$\ce{NaOAc + HCl -> HOAc + NaCl} \tag{1}$$
Suppose you added $x$ volume of $\ce{NaOAc}$ and $y$ volume of $\ce{HCl}$ to make the $\pu{1.0L}$ of buffer ($\mathrm{pH} = 4.43$). Thus if units of $x$ and $y$ are $\pu{L}$, then:
$$x + y = 1 \tag {2}$$
And, after acid-base reaction, according to the chemical equation given above and equation $(2)$, remaining amounts of $\ce{NaOAc}$ in $\pu{mol}$ is $(0.3x-0.2y)=\left(0.3x-0.2(1-x)\right)= (0.5x-0.2)$ and amounts of $\ce{HOAc}$ formed in $\pu{mol}$ is $0.2y= 0.2(1-x)=(0.2-0.2x)$. Thus, since total volume of buffer is $\pu{1.0 L}$, respective concentrations of $\ce{NaOAc}$ and $\ce{HOAc}$ in the solution are as follows:
$$\ce{[NaOAc]} = (0.5x-0.2)/1.0=(0.5x-0.2) \text{ and } \ce{[HOAc]} = (0.2-0.2x)/1.0 = 0.2-0.2x$$
Now you can substitute these values to Henderson-Hasselbalch equation:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{\ce{[NaOAc]}}{\ce{[HOAc]}}$$ 
$$4.43 = 4.76 + \log \frac{(0.5x-0.2)}{(0.2-0.2x)}$$ 
Thus,
$$ \log \frac{(0.5x-0.2)}{(0.2-0.2x)}= 4.43 - 4.76= -0.33$$
$$ \frac{(0.5x-0.2)}{(0.2-0.2x)}= 10^{-0.33} = 0.468$$
Once you solved for $x$, you would get, $x=0.495$ and $y=1.0-0.495=0.505$ in $\pu{L}$, respectively (or $495$ and $\pu{505 mL}$). Thus, I'd say the given answers for your question is incorrect.

