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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Equilibrium
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Which of the following solutions will have $pH$ of 4.74?

(a) 100 mL of 1 M $CH_3COOH$ ($pK_a = 4.74$) at the equivalence point using 1 M NaOH

(b) 50 mL of 1 M $CH_3COONa$ + 25 mL of 1 M HCl

(c) 50 mL of 1 M $CH_3COOH$ + 25 mL of 1 M NaOH

(d) Both (b) and (c)

Can you answer this question?
 
 

1 Answer

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Answer: 50 mL of 1 M $CH_3COONa$ + 25 mL of 1 M HCl and 50 mL of 1 M $CH_3COOH$ + 25 mL of 1 M NaOH both
 
In acid base titration, at equivalence point $pH = pK_{in}$
 
50 mL of 1 M $CH_3COONa$ + 25 mL of 1 M HCl = 25 mL of 1 M $CH_3COOH$ + 25 mL of 1 M NaCl
This is because strong acid HCl will displace weak acid, $CH_3COOH$
 
In case of an acid buffer solution when
[salt] = [acid]
$pH = pK_a = 4.74$
$\therefore$ Choice (B) is correct.
 
At mid point of a titration of a weak acid and strong base
$pH = pK_a = 4.74$
$\therefore$ Choice (C) is also correct

 

answered Nov 29, 2013 by mosymeow_1
 

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