ExamBro
ExamBro
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 

  1. A \(4: 5\)
  2. B \(1 : 10\)
  3. C \(10 : 1\)
  4. D \(5 : 4\)
Verified Solution

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, \(pH = 6,\,[{H^ + }] = 1 \times {10^{ - 6}}\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app