$A$ buffer solution is prepared by mixing $0.01 \ M$ $HCN$ and $0.02 \ M$ $NaCN$. If $K_{a}$ for $HCN$ is $6.6 \times 10^{-10}$,what is the concentration of $H^{+}$ ions in the solution?
- A$3.3 \times 10^{-6} \ M$
- B$3.3 \times 10^{-10} \ M$
- C$1.32 \times 10^{-6} \ M$
- D$1.32 \times 10^{-10} \ M$
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Similar Questions
All the given solutions have the same concentration. Mixing equal volumes of which of the following will produce a buffer solution?
$A = NH_4Cl$; $B = CH_3COONa$; $C = NH_4OH$; $D = CH_3COOH$
$A = NH_4Cl$; $B = CH_3COONa$; $C = NH_4OH$; $D = CH_3COOH$
Medium
View SolutionCalculate the amount of $(NH_4)_2SO_4$ in grams which must be added to $500 \ mL$ of $0.200 \ M \ NH_3$ to yield a solution with $pH = 9.35$ ($K_b$ for $NH_3 = 1.78 \times 10^{-5}$).
Medium
View Solution$A$ 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; $\log 2=0.30$ ]
[Given : $pK_{a1}$ and $pK_{a2}$ of $H_2CO_3$ are $6.37$ and $10.32$,respectively; $\log 2=0.30$ ]
MediumIIT 2022
View SolutionDerive the Henderson-Hasselbalch equation.
Difficult
View SolutionAssertion : In a titration of weak acid and $NaOH$,the $pH$ at half equivalence point is $pK_a$.
Reason : At half equivalence point,it forms an acidic buffer and the buffer capacity is maximum where $[acid] = [salt]$.
Reason : At half equivalence point,it forms an acidic buffer and the buffer capacity is maximum where $[acid] = [salt]$.
MediumAIIMS 2009
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