(A) For an acidic buffer,the Henderson-Hasselbalch equation is given by: $pH = pK_{a} + \log \frac{[\text{Salt}]}{[\text{Acid}]}$ Given that the conce

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(A) For an acidic buffer,the Henderson-Hasselbalch equation is given by: $pH = pK_{a} + \log \frac{[\text{Salt}]}{[\text{Acid}]}$ Given that the concentrations of the weak acid and its salt are equal,i.e.,$[\text{Salt}] = [\text{Acid}]$. Therefore,the equation simplifies to: $pH = pK_{a} = -\log K_{a}$ Given $K_{a} = 1.8 \times 10^{-5}$. $pH = -\log (1.8 \times 10^{-5}) = -(\log 1.8 + \log 10^{-5})$ $pH = -(\log 1.8 - 5) = 5 - \log 1.8$ Using $\log 1.8 \approx 0.2553$: $pH = 5 - 0.2553 = 4.7447$

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Buffer solution

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$A$ buffer solution contains equal concentrations of weak acid and its salt with a strong base. Calculate the $pH$ of the buffer solution if the dissociation constant of the weak acid is $1.8 \times 10^{-5}$.

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A
$4.7447$
B
$5.142$
C
$5.8496$
D
$4.0128$

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6-2.Equilibrium-II (Ionic Equilibrium)

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$A$ buffer solution contains equal concentrations of weak acid and its salt with a strong base. Calculate the $pH$ of the buffer solution if the dissociation constant of the weak acid is $1.8 \times 10^{-5}$.

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