JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
Addition of sodium hydroxide solution to a weak acid \((HA)\) results in a buffer of \(pH\, 6\). lf ionisation constant of \(HA\) is \(10^{-5}\) , the ratio of salt to acid concentration in the buffer solution will be
- A \(4: 5\)
- B \(1 : 10\)
- C \(10 : 1\)
- D \(5 : 4\)
Answer & Solution
Correct Answer
(C) \(10 : 1\)
Step-by-step Solution
Detailed explanation
\(HA \leftrightarrow {H^ + } + {A^ - }\) (Unionized, weak acid and common ion effect) \(HA + NaOH \to NaA + {H_2}O\) \(NaA \to N{a^ + } + {A^ - }\) (ionized) \({K_a} = \frac{{[{H^ + }][{A^ - }]}}{{[HA]}}\) Given, \(pH = 6,\,[{H^ + }] = 1 \times {10^{ - 6}}\)…
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