MCQ
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
    $1 : 10$
  • $10 : 1$
  • D
    $5 : 4$

Answer

Correct option: C.
$10 : 1$
c
$HA \leftrightarrow {H^ + } + {A^ - }$

(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,  $pH = 6,\,[{H^ + }] = 1 \times {10^{ - 6}}$

$[{H^ + }] = \frac{{{K_a}[Acid]}}{{[Salt]}}$

$\frac{{[Salt]}}{{[Acid]}} = \frac{{{K_a}}}{{[{H^ + }]}} = \frac{{{{10}^{ - 5}}}}{{{{10}^{ - 6}}}} = 10:1$

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