Answer
$$0.14 \space g \space NaOH$$
Work Step by Step
$$H_2CO_3 (aq) +NaOH(aq) --\gt HC{O_3}^-(aq) + H_2O(l) $$ \begin{matrix} & [H_2C{O_3}] & [NaOH] & [HCO{_3}^-] \\ Initial & 0.0012& x & 0.024 \\ Change & -x & -x & +x\\ Equil & 0.0012 -x & 0 & 0.024 + x \end{matrix}
$$pH = 6.1 + log\frac{[Base]}{[Acid]}$$ $$7.8 = 6.1 + log\frac{0.024 \space + x}{0.0012 \space - x}$$ $$7.8 - 6.1= + log\frac{0.024 \space + x}{0.0012 \space - x}$$ $$1.7= log\frac{0.024 \space + x}{0.0012 \space - x}$$ $$10^{1.7}=\frac{0.024 \space + x}{0.0012 \space - x}$$ $$(0.0012)10^{1.7} - 10^{1.7}x=0.024 \space + x$$ $$(0.0012)10^{1.7} - 0.024 = x + 10^{1.7}x$$ $$(0.0012)10^{1.7} - 0.024 = x (1 + 10^{1.7})$$ $$\frac{(0.0012)10^{1.7} - 0.024}{1 + 10^{1.7}} = x $$ $$x = 7.0703 \times 10^{-4}$$
1. Since the molarity of that solution is equal to $ 0.00070703 \space M \space NaOH$:
$1 \space L \space solution = 0.00070703 \space mole \space NaOH$
$\frac{1 \space L \space solution}{ 0.00070703 \space mole \space NaOH} $ and $\frac{ 0.00070703 \space mole \space NaOH}{1 \space L \space solution}$
2. Determine the molar mass of this compound (NaOH), and setup the conversion factors:
Molar mass :
$Na: 22.99g $
$O: 16g $
$H: 1.008g$
22.99g + 16g + 1.008g = 39.998g
$ \frac{1 \space mole \space (NaOH)}{ 39.998 \space g \space (NaOH)}$ and $ \frac{ 39.998 \space g \space (NaOH)}{1 \space mole \space (NaOH)}$
3. Use the conversion factors to calculate the mass of solute in $ 5$ L of that solution:
$ 5.0 \space L \space solution \times \frac{ 0.00070703 \space mole \space NaOH}{1 \space L \space solution} \times \frac{ 39.998 \space g \space NaOH}{1 \space mole \space NaOH} = 0.14 \space g \space NaOH$
