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(A) The initial moles are $0.01 \ mol$ $(10 \ mmol)$ each of $H_2CO_3$,$NaHCO_3$,$Na_2CO_3$,and $NaOH$.Step $1$: Reaction between $H_2CO_3$ and $NaOH$

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$10.1$

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(A) The initial moles are $0.01 \ mol$ $(10 \ mmol)$ each of $H_2CO_3$,$NaHCO_3$,$Na_2CO_3$,and $NaOH$.Step $1$: Reaction between $H_2CO_3$ and $NaOH$:$H_2CO_3 + NaOH \longrightarrow NaHCO_3 + H_2O$Initial: $10 \ mmol, 10 \ mmol, 10 \ mmol$Final: $0 \ mmol, 0 \ mmol, 20 \ mmol$Step $2$: The final mixture contains $20 \ mmol$ of $NaHCO_3$ (from initial $10 \ mmol$ + $10 \ mmol$ produced) and $10 \ mmol$ of $Na_2CO_3$.Step $3$: This forms a buffer solution of $NaHCO_3$ (acid) and $Na_2CO_3$ (salt).Using the Henderson-Hasselbalch equation:$pH = pK_{a2} + \log \left( \frac{[Na_2CO_3]}{[NaHCO_3]} \right)$$pH = 10.32 + \log \left( \frac{10}{20} \right)$$pH = 10.32 - \log 2$$pH = 10.32 - 0.30 = 10.02$Rounding to the nearest option,the answer is $10.1$.

$A$ solution is prepared by mixing $0.01 \ mol$ each of $H_2CO_3$,$NaHCO_3$,$Na_2CO_3$,and $NaOH$ in $100 \ mL$ of water. The $pH$ of the resulting solution is. . . . . . .
[Given : $pK_{a1}$ and $pK_{a2}$ of $H_2CO_3$ are $6.37$ and $10.32$,respectively; $\log 2=0.30$ ]

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$A$ solution is prepared by mixing $0.01 \ mol$ each of $H_2CO_3$,$NaHCO_3$,$Na_2CO_3$,and $NaOH$ in $100 \ mL$ of water. The $pH$ of the resulting solution is. . . . . . .[Given : $pK_{a1}$ and $pK_{a2}$ of $H_2CO_3$ are $6.37$ and $10.32$,respectively; $\log 2=0.30$ ]