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25.0 mL of 0.1M NaOh is titred with 0.1M...

`25.0 mL` of `0.1M NaOh` is titred with `0.1M HC1`. Calculate `pH` when:
i. `20mL` ii. `24mL` of acid is added.

Text Solution

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The correct Answer is:
A, B, C
Mmoles of `NaOH = 25 xx 0.1 = 2.5`
a. `20mL` of `0.1 M HC1` is added.
mmol of `HC1 = 20 xx 0.1 = 2`
mmol of `NaOH` left
`= 2.5 - 2.0 = 0.5` mmol
Volume `=25 + 20 = 45mL`
`[overset(Theta)OH] = (0.5m mol)/(45mL) ~~ 0.01 = 10^(-2)M`
`pOH = 2. pH = 12`
b. `24mL of 0.1 M HC1` is added.
mmol of `HC1 = 24 xx 0.1 = 24`
mmol of `NaOH` left `= 2.5 - 2.4 = 0.1`
`[overset(Theta)OH] = (0.1)/((25 +24)) = 0.002 = 2 xx10^(-3)M`
`pOH =- log [2 xx 10^(-3)] =- 0.3 +3 = 2.7`
`pH = 14 - 2.7 = 11.3`
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