logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Equilibrium
0 votes

$pH$ of a mixture of 1 M benzoic acid ($pK_a$ = 4.20) and 1 M solution benzoate is 4.5. In 300 mL of buffer solution, volume of benzoic acid solution is

(a) 200 mL

(b) 150 mL

(c) 100 mL

(d) 50 mL

Can you answer this question?
 
 

1 Answer

0 votes
Answer = 100 mL
 
$pH = pK_a + log \frac{[Salt]}{[Acid]}$
 
Putting the given values, $4.5 = 4.2 + log \frac{[Salt]}{[Acid]}$
$0.3 = log \frac{[Salt]}{[Acid]}$
$\frac{[Salt]}{[Acid]}$ = antilog 0.3 = 2
 
Let v mL of benzoic acid is mixed with (300-v) mL of 1M sodium benzoate.
$\therefore \frac{300-v}{v} = 2$
$\Rightarrow 300 - v = 2 v$
v = 100 mL

 

answered Nov 29, 2013 by mosymeow_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...