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`        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 moleNumber 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 molesHence 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 = 2.301Regards Askiitian member If you like this  answer please encourage us by approving this answer.
```
one year ago
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