HOW TO SOLVE THIS ACCODING TO THIS GUSTION

First, determine the molar mass of $\ce{NaOH}$:
Molar mass of $\ce{NaOH} = 22.99\ (\text{Na}) + 16.00\ (\text{O}) + 1.01\ (\text{H}) = 40.00\ \text{g/mol}$

Next, calculate the moles of $\ce{NaOH}$ present:
$\text{Moles of NaOH} = \frac{\text{Mass of NaOH}}{\text{Molar mass of NaOH}} = \frac{5.00\ \text{g}}{40.00\ \text{g/mol}} = 0.125\ \text{mol}$

Since the volume of the solution is $1.00\ \text{L}$, the molarity of $\ce{NaOH}$ is:
$[\ce{NaOH}] = \frac{0.125\ \text{mol}}{1.00\ \text{L}} = 0.125\ \text{M}$

As $\ce{NaOH}$ is a strong base, it dissociates completely, so the concentration of $\ce{OH-}$ from $\ce{NaOH}$ is:
$[\ce{OH-}]_{\text{NaOH}} = 0.125\ \text{M}$

First, determine the molar mass of $\ce{Na2CO3}$:
Molar mass of $\ce{Na2CO3} = (2 \times 22.99)\ (\text{Na}) + 12.01\ (\text{C}) + (3 \times 16.00)\ (\text{O}) = 45.98 + 12.01 + 48.00 = 105.99\ \text{g/mol}$

Next, calculate the moles of $\ce{Na2CO3}$ present:
$\text{Moles of Na2CO3} = \frac{\text{Mass of Na2CO3}}{\text{Molar mass of Na2CO3}} = \frac{5.00\ \text{g}}{105.99\ \text{g/mol}} = 0.047174\ \text{mol}$

Since the volume of the solution is $1.00\ \text{L}$, the molarity of $\ce{Na2CO3}$ is:
$[\ce{Na2CO3}] = \frac{0.047174\ \text{mol}}{1.00\ \text{L}} = 0.047174\ \text{M}$

$\ce{Na2CO3}$ dissociates completely into $\ce{2Na+}$ and $\ce{CO3^2-}$, so the initial concentration of carbonate ion is:
$[\ce{CO3^2-}]_0 = 0.047174\ \text{M}$

The carbonate ion acts as a weak base in water:
$\ce{CO3^2- (aq) + H2O (l) <=> HCO3- (aq) + OH- (aq)}$
The $K_b$ for this reaction is related to the $K_a$ of its conjugate acid, $\ce{HCO3-}$ ($K_{a2}$ for $\ce{H2CO3}$).
Given $K_{a2}$ for $\ce{H2CO3} = 5.6 \times 10^{-11}$.
$K_b = \frac{K_w}{K_{a2}} = \frac{1.0 \times 10^{-14}}{5.6 \times 10^{-11}} = 1.7857 \times 10^{-4}$

We set up an ICE table for the hydrolysis of $\ce{CO3^2-}$:
$\ce{CO3^2- (aq) + H2O (l) <=> HCO3- (aq) + OH- (aq)}$
Initial: $0.047174\ \text{M}$ $0\ \text{M}$ $0.125\ \text{M}$ (from $\ce{NaOH}$)
Change: $-x$ $+x$ $+x$
Equilibrium: $0.047174 - x$ $x$ $0.125 + x$

The equilibrium expression is:
$K_b = \frac{[\ce{HCO3-}][\ce{OH-}]}{[\ce{CO3^2-}]} = \frac{(x)(0.125 + x)}{0.047174 - x}$

Since $K_b$ is small and the initial $[\ce{OH-}]$ is relatively large, we can assume $x$ is much smaller than 0.047174 and 0.125.
So, $1.7857 \times 10^{-4} \approx \frac{(x)(0.125)}{0.047174}$
$x = \frac{(1.7857 \times 10^{-4}) \times 0.047174}{0.125}$
$x = 6.739 \times 10^{-5}\ \text{M}$

This value of $x$ is indeed much smaller than 0.047174 and 0.125, so the approximation is valid.
The contribution of $\ce{OH-}$ from $\ce{CO3^2-}$ is $6.739 \times 10^{-5}\ \text{M}$.

Total [\ce{OH-}] = [\ce{OH-}]_{\text{NaOH}} + [\ce{OH-}]_{\text{from CO3^2-}}
Total $[\ce{OH-}] = 0.125\ \text{M} + 6.739 \times 10^{-5}\ \text{M}$
Total $[\ce{OH-}] = 0.12506739\ \text{M}$

$\text{pOH} = -\log[\text{Total } \ce{OH-}]$
$\text{pOH} = -\log(0.12506739)$
$\text{pOH} = 0.9028$

Finally, calculate the pH using the relationship $\text{pH} + \text{pOH} = 14.00$:
$\text{pH} = 14.00 - \text{pOH}$
$\text{pH} = 14.00 - 0.9028$
$\text{pH} = 13.0972$

Rounding to two decimal places, consistent with the precision of the given $K_a$ value:
$\text{pH} = 13.10$

The final answer is 13.10.