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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Equilibrium
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Degree of dissociation of an acid HCl is 95%. 0.192 g of the acid is present in 0.5 L of solution. The pH of the solution is

(a) 2

(b) 1

(c) 3

(d) 4

Can you answer this question?
 
 

1 Answer

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Toolbox:
  • $\rho H = - log [H^+]$
 
Answer: 2
 
No. of moles of $HCl = \frac{0.192}{36.5}$ = 0.0053
Volume = 0.5 litre
 
Molarity of $HCL = \frac{0.0053}{0.5} = 1.06 \times 10^{-2}$ M
 
$[H^+]$ in $HCL$ solution = $1.06 \times 10^{-2} \times \frac{95}{100} = 100.7 \times 10^{-4} \approx 1 \times 10^{-2} $
 
Thus, $\rho H = - log [H^+] = - log (1 \times 10^{-2}) = 2$

 

answered Nov 23, 2013 by mosymeow_1
 

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