A solution is prepared by mixing 0.01 mol each of $$H_2CO_3$$, NaHCO$$_3$$, Na$$_2$$CO$$_3$$, and NaOH in 100 mL of water. pH of the resulting solution is ______.

[Given : $$pK_{a1}$$ and $$pK_{a2}$$ of $$H_2CO_3$$ are 6.37 and 10.32, respectively ; log 2 = 0.30]

Initially, each solute furnishes $$0.01\text{ mol}$$ in a total volume of $$100\text{ mL}=0.1\text{ L}$$, so their starting concentrations are $$0.1\text{ M}$$ each:

$$H_2CO_3\;(0.1\text{ M}),\;HCO_3^- \;(0.1\text{ M}),\;CO_3^{2-}\;(0.1\text{ M}),\;OH^-\;(0.1\text{ M})$$

Step 1: Immediate neutralisation by the strong base
The strong base $$OH^-$$ will first de-protonate the strongest acid present, $$H_2CO_3$$:

$$H_2CO_3 + OH^- \;\longrightarrow\; HCO_3^- + H_2O$$

Mole balance (in 0.1 L): both $$H_2CO_3$$ and $$OH^-$$ start with $$0.01\text{ mol}$$, so they completely consume each other.

After this reaction:

$$H_2CO_3 : 0$$    $$OH^- : 0$$    additional $$HCO_3^- : +0.01\text{ mol}$$

Step 2: Concentrations after neutralisation
Total moles now present:

$$HCO_3^- : 0.01 + 0.01 = 0.02\text{ mol}$$
$$CO_3^{2-} : 0.01\text{ mol}$$

Convert to concentrations (divide by $$0.1\text{ L}$$):

$$[HCO_3^-] = \frac{0.02}{0.1} = 0.20\text{ M}$$
$$[CO_3^{2-}] = \frac{0.01}{0.1} = 0.10\text{ M}$$

Step 3: Identify the buffer pair
The remaining species constitute the conjugate acid-base pair of the second dissociation of carbonic acid:

$$HCO_3^- \;\rightleftharpoons\; H^+ + CO_3^{2-} \quad\quad (pK_{a2} = 10.32)$$

Step 4: Apply the Henderson-Hasselbalch equation

$$\text{pH} = pK_{a2} + \log\!\left(\frac{[CO_3^{2-}]}{[HCO_3^-]}\right)$$

Substitute the concentrations:

$$\text{pH} = 10.32 + \log\!\left(\frac{0.10}{0.20}\right) = 10.32 + \log(0.50)$$

Using $$\log 2 = 0.30 \;,$$ so $$\log(0.50) = -\log 2 = -0.30$$, we get

$$\text{pH} = 10.32 - 0.30 = 10.02$$

Hence, the pH of the resulting solution is 10.02.