To calculate the pH of a solution resulting from the titration of carbonic acid with sodium hydroxide, we start by determining the equivalence volume needed to reach the first equivalence point. This is done using the formula:

m_{\text{acid}} \times V_{\text{acid}} = m_{\text{base}} \times V_{\text{base}}

Given that both the carbonic acid and sodium hydroxide have a concentration of 0.250 M, we can set up the equation:

0.250 \, \text{mol/L} \times 100 \, \text{mL} = 0.250 \, \text{mol/L} \times V_{\text{base}}

Solving for the volume of the base needed to reach the first equivalence point gives us:

V_{\text{base}} = 100 \, \text{mL}

Next, we find the volume of sodium hydroxide that actually reacted by subtracting the equivalence volume from the total volume added:

V_{\text{reacting}} = 120 \, \text{mL} - 100 \, \text{mL} = 20 \, \text{mL}

Since we have exceeded the first equivalence point, we recognize that we are now dealing with the intermediate form of carbonic acid, which is bicarbonate (HCO₃^-). The second equivalence point would require an additional 100 mL of NaOH, totaling 200 mL to reach it, but we are only at 120 mL.

To analyze the system, we set up an ICF (Initial, Change, Final) chart. Initially, we have:

- Moles of bicarbonate (intermediate form): 0.025 \, \text{moles} (calculated from 0.250 M and 100 mL)

- Moles of sodium hydroxide reacting: 0.005 \, \text{moles} (from 20 mL of 0.250 M NaOH)

- Moles of conjugate base (carbonate ion, CO₃^2-): 0

After the reaction, we have:

- Moles of bicarbonate remaining: 0.020 \, \text{moles} (0.025 - 0.005)

- Moles of sodium hydroxide remaining: 0

- Moles of carbonate formed: 0.005 \, \text{moles}

At this point, we have a buffer solution consisting of bicarbonate and carbonate ions. To find the pH of this buffer, we use the Henderson-Hasselbalch equation:

pH = pK_a + \log\left(\frac{[\text{base}]}{[\text{acid}]}\right)

Here, we use pK_a2, which is calculated as:

pK_a2 = -\log(5.6 \times 10^{-11})

Substituting the values into the equation, we have:

pH = pK_a2 + \log\left(\frac{0.005 \, \text{moles}}{0.020 \, \text{moles}}\right)

Calculating this gives us a final pH of approximately 9.65 for the solution after the titration.