NEET · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
At 298 K, a certain buffer solution contains equal concentrations of \(X^{-}\)and \(H X, K_b\) for \(X^{-}\)is \(10^{-10}\). What is the pH of this buffer solution ?
- A 2
- B 4
- C 6
- D 10
Answer & Solution
Correct Answer
(B) 4
Step-by-step Solution
Detailed explanation
(B) Given \(K_b\) for \(X^{-}=10^{-10}\).
For a conjugate acid-base pair at \(298 K, K_a \times K_b=K_w=10^{-14}\).
\(K_a=\frac{10^{-14}}{10^{-10}}=10^{-4}\)
The \(p K_a\) of the weak acid \(H X\) is:
\(p K_a=-\log \left(K_a\right)=-\log \left(10^{-4}\right)=4\)
Using the Henderson-Hasselbalch equation for an acidic buffer:
\(pH=p K_a+\log \left(\frac{\left[X^{-}\right]}{[H X]}\right)\)
Since the concentrations of \(X^{-}\)and \(H X\) are equal, \(\left[X^{-}\right]=[H X]\).
\(pH=4+\log (1)=4+0=4\)
For a conjugate acid-base pair at \(298 K, K_a \times K_b=K_w=10^{-14}\).
\(K_a=\frac{10^{-14}}{10^{-10}}=10^{-4}\)
The \(p K_a\) of the weak acid \(H X\) is:
\(p K_a=-\log \left(K_a\right)=-\log \left(10^{-4}\right)=4\)
Using the Henderson-Hasselbalch equation for an acidic buffer:
\(pH=p K_a+\log \left(\frac{\left[X^{-}\right]}{[H X]}\right)\)
Since the concentrations of \(X^{-}\)and \(H X\) are equal, \(\left[X^{-}\right]=[H X]\).
\(pH=4+\log (1)=4+0=4\)
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