A solution is prepared by mixing 0.01 mol ea | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $NaOH$

Vedclass · Accepted Answer

$10.1$

Vedclass · Answer

(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; $\log 2=0.30$ ]

Similar Questions

$50 \ mL$ of $0.02 \ M$ $NaHSO_4$ is mixed with $50 \ mL$ of $0.02 \ M$ $Na_2SO_4$. Calculate the $pH$ of the resulting solution. $[pK_{a2} (H_2SO_4) = 2]$

In a buffer solution containing equal concentration of $B^{-}$ and $HB,$ the $K_b$ for $B^{-}$ is $10^{-10}.$ The $pH$ of the buffer solution is:

The buffer system which helps to maintain the $pH$ of blood between $7.26$ to $7.42$ is:

$A$ buffer solution with $pH = 9$ is prepared by mixing $NH_4Cl$ and $NH_4OH$. Calculate the number of moles of $NH_4Cl$ dissolved in $1.0 \, L$ of $1.0 \, M \, NH_4OH$ solution. (Given: $K_b(NH_4OH) = 1.8 \times 10^{-5}$)

In a mixture of acetic acid and sodium acetate,the ratio of the concentration of the salt to the acid is increased ten times. Then the $pH$ of the solution:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$A$ 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; $\log 2=0.30$ ]