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Byju's Answer

Standard XII

Chemistry

Proteins

In the titrat...

Question

In the titration of a solution of a weak acid HX with NaOH, the pH is 5.8 after 10 ml of NaOH solution has been added to it, and it is 6.4 after 20 ml of NaOH has been added to it. Find the $p K_{a}$ of HX? (Take $10^{0.6} = 4.0$ and $l o g 2 = 0.3$ )

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Solution

Let $M_{1}$ and $M_{2}$ be the molarities of Acid HX and the base NaOH respectively and let 'V' be the volume of acid taken, now substituting the given data in Henderson's equation we get:
$p H = p K_{a} + log \frac{[salt]}{[Acid]}$
$5.8 = p K_{a} + log [\frac{(10 m l) M_{2}}{V M_{1} - (10 m l) M_{2}}] 6.4 = p K_{a} + log [\frac{(20 m l) M_{2}}{V M_{1} - (20 m l) M_{2}}]$
Subtracting the above two equations we get:
$0.6 = log [\frac{(20 m l) M_{2} . V M_{1} - (10 m l) M_{2}}{V M_{1} - (20 m l) M_{2} . (10 m l) M_{2}}] o r, \frac{2 [V M_{1} - (10 m l) M_{2}]}{V M_{1} - (20 m l) M_{2}} = 4$
(as given $10^{0.6} = 4$ )
or, $\frac{V M_{1}}{M_{2}} = \frac{60 m l}{2} = 30 m l$
Substituting this in either Eq (1)
or Eq. (2) ; we get $p K_{a} = 5.8 - log (\frac{10}{30 - 10})$
$= 5.8 + 0.30 = 6.1$

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In the titration of a solution of a weak acid HX with NaOH, the pH is 5.8 after 10 ml of NaOH solution has been added to it, and it is 6.4 after 20 ml of NaOH has been added to it. Find the pKa of HX? (Take 100.6=4.0 and log2=0.3)

In the titration of a solution of a weak acid HX with NaOH, the pH is 5.8 after 10 ml of NaOH solution has been added to it, and it is 6.4 after 20 ml of NaOH has been added to it. Find the $p K_{a}$ of HX? (Take $10^{0.6} = 4.0$ and $l o g 2 = 0.3$ )