Answer
$$0.3 \space g \space HCl$$
Work Step by Step
$$HC{O_3}^-(aq) + HCl(l) --\gt H_2CO_3 (aq) + Cl^-(aq) $$ \begin{matrix} & [HC{O_3}^-] & [HCl] & [H_2CO_3] \\ Initial & 0.024& x & 0.0012 \\ Change & -x & -x & +x\\ Equil & 0.024 -x & 0 & 0.0012 + x \end{matrix}
$$pH = 6.1 + log\frac{[Base]}{[Acid]}$$ $$7.0 = 6.1 + log\frac{0.024 \space - x}{0.0012 \space + x}$$ $$0.9 = log\frac{0.024 \space - x}{0.0012 \space + x}$$ $$10^{0.9} = \frac{0.024 \space - x}{0.0012 \space + x}$$ $$(0.0012)10^{0.9} + 10^{0.9}x = 0.024 \space - x$$ $$(0.0012)10^{0.9} - 0.024 = - x - 10^{0.9}x$$ $$\frac{(0.0012)10^{0.9} - 0.024}{-1 - 10^{0.9}} = x $$ $$x = 1.6178 \times 10^{-3}$$
1. Since the molarity of that solution is equal to $ 0.0016178 \space M \space HCl$:
$1 \space L \space solution = 0.0016178 \space mole \space HCl$
$\frac{1 \space L \space solution}{ 0.0016178 \space mole \space HCl} $ and $\frac{ 0.0016178 \space mole \space HCl}{1 \space L \space solution}$
2. Determine the molar mass of this compound (HCl), and setup the conversion factors:
Molar mass :
$H: 1.008g $
$Cl: 35.45g$
1.008g + 35.45g = 36.46g
$ \frac{1 \space mole \space (HCl)}{ 36.46 \space g \space (HCl)}$ and $ \frac{ 36.46 \space g \space (HCl)}{1 \space mole \space (HCl)}$
3. Use the conversion factors to calculate the mass of solute in $ 5.0$ L of that solution:
$ 5.0 \space L \space solution \times \frac{ 0.0016178 \space mole \space HCl}{1 \space L \space solution} \times \frac{ 36.46 \space g \space HCl}{1 \space mole \space HCl} = 0.295 \space g \space HCl$
Since we have used a lot of approximations, the result should have a low number of significant figures. I am going to use exactly 1.
$$0.3 \space g \space HCl$$
