Grade 12th passPhysical Chemistry
Calculate the pH of a solution of 50ml 0.1N NaOH and 100ml 0.1N CH3COOH.
Calculate the pH of a solution of 50ml 0.1N NaOH and 100ml 0.1N CH3COOH.
1 Answer
Sahani Kumar
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
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