The $pH$ of a sodium acetate buffer solution is given by the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$. For acetic acid,if $[Salt] = [Acid] = 0.1 \ M$,then the $pH$ of the solution is: $[K_a = 1.8 \times 10^{-5}]$
- A$7$
- B$4.74$
- C$5.3$
- D$1.4$
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$20 \ mL$ of $0.1 \ M \ H_2SO_4$ is added to $30 \ mL$ of $0.2 \ M \ NH_4OH$ solution. The $pH$ of the resultant mixture is [$pK_b$ of $NH_4OH = 4.7$].
DifficultJEE MAIN 2019
View Solution$100 \ mL$ of $0.1 \ M$ $HA$ (weak acid) and $100 \ mL$ of $0.2 \ M$ $NaA$ are mixed. What is the $pH$ of the resultant solution? ($K_{a}$ of $HA$ is $10^{-5}$; $\log 2 = 0.3$)
$A$ solution is prepared by mixing $10 \ mL$ of $1.0 \ M$ acetic acid and $20 \ mL$ of $0.5 \ M$ sodium acetate and diluted to $100 \ mL$. If the $pK_{a}$ of acetic acid is $4.76$,then the $pH$ of the solution is
The $pH$ of a buffer solution made by mixing $25 \ mL$ of $0.02 \ M$ $NH_4OH$ and $25 \ mL$ of $0.2 \ M$ $NH_4Cl$ at $25^{\circ}C$ is ($pK_b$ of $NH_4OH = 4.8$).
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View SolutionWhich of the following does not act as a buffer solution?
Medium
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