A solution is prepared by mixing 0.01 mol each of H 2 CO 3, NaHCO 3, Na 2 CO 3, and NaOH in 100 mL of water. pH of the resulting solution is [Given: p K a 1 and p K a 2 of H 2 CO 3 are 6.37 and 10.32, respectively ; log 2=0.30 ]

Question

A solution is prepared by mixing 0.01 mol each of H 2 CO 3, NaHCO 3, Na 2 CO 3, and NaOH in 100 mL of water. pH of the resulting solution is [Given: p K a 1 and p K a 2 of H 2 CO 3 are 6.37 and 10.32, respectively ; log 2=0.30 ] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/efbac46864f25565f5fd75b26a0c6a90-.png" /> Final mixture has 20 milli moles NaHCO 3 and 10 milli moles Na 2 CO 3 pH = pKa 2+ log ( text Salt / text Acid ) pH = pKa 2+ log ((10/20)) [Buffer: Na 2 CO 3+ NaHCO 3 ] =10.32- log 2=10.02

Q. A solution is prepared by mixing $0.01 m o l$ each of $H_{2} C O_{3}, N a H C O_{3}, N a_{2} C O_{3}$ , and $N a O H$ in $100 m L$ of water. $p H$ of the resulting solution is ___
[Given : $p K_{a 1}$ and $p K_{a 2}$ of $H_{2} C O_{3}$ are $6.37$ and $10.32$ , respectively $; lo g 2 = 0.30$ ]

Final mixture has 20 milli moles $N a H C O_{3}$ and 10 milli moles $N a_{2} C O_{3}$
$p H = p K a_{2} + lo g \frac{Salt}{Acid}$
$p H = p K a_{2} + lo g (\frac{10}{20})$
[Buffer : $N a_{2} C O_{3} + N a H C O_{3}$ ]
$= 10.32 - lo g 2 = 10.02$

Q. A solution is prepared by mixing 0.01mol each of H2​CO3​,NaHCO3​,Na2​CO3​, and NaOH in 100mL of water. pH of the resulting solution is ___ [Given : pKa1​ and pKa2​ of H2​CO3​ are 6.37 and 10.32, respectively ;log2=0.30 ]

Final mixture has 20 milli moles NaHCO3​ and 10 milli moles Na2​CO3​ pH=pKa2​+log Acid Salt ​ pH=pKa2​+log(2010​) [Buffer : Na2​CO3​+NaHCO3​ ] =10.32−log2=10.02

Q. A solution is prepared by mixing $0.01 m o l$ each of $H_{2} C O_{3}, N a H C O_{3}, N a_{2} C O_{3}$ , and $N a O H$ in $100 m L$ of water. $p H$ of the resulting solution is ___
[Given : $p K_{a 1}$ and $p K_{a 2}$ of $H_{2} C O_{3}$ are $6.37$ and $10.32$ , respectively $; lo g 2 = 0.30$ ]

Final mixture has 20 milli moles $N a H C O_{3}$ and 10 milli moles $N a_{2} C O_{3}$
$p H = p K a_{2} + lo g \frac{Salt}{Acid}$
$p H = p K a_{2} + lo g (\frac{10}{20})$
[Buffer : $N a_{2} C O_{3} + N a H C O_{3}$ ]
$= 10.32 - lo g 2 = 10.02$