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Grade: 12th pass
        
Calculate the pH of a solution of 50ml 0.1N NaOH and 100ml 0.1N CH3COOH.
one year ago

Answers : (1)

Sahani Kumar
99 Points
							
Number of moles of NaOH = (0.1 * 50)/(1000)
                                                = 5 * 10^-3 mole
Number of moles of CH3COOH= (0.1 * 100)/(1000)
                                                       = 10 * 10^-3 moles 
But 5 * 10^-3 moles of NaOH will neutralize 
5 * 10^-3 moles of CH3COOH 
Hence number of moles of CH3COOH left after the neutralization= ((10 *10^-3) – (5 *10^-3))
                         = 5 * 10^-3 moles
Hence 5 * 10^-3 moles of CH3COOH will produce 
5 * 10^-3 moles of H +.
Now 
We know that 
– pH = log (H+)
– pH = log (5 * 10^-3)
– pH = log(5) + log(10^-3)
– pH =0.699 – 3
   pH = 3 – 0.699 = 2.301
Regards 
Askiitian member 
If you like this  answer please encourage us by approving this answer. 
 
one year ago
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