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25.0mL ofO.1 M NaOH is titrated with 0.1...

25.0mL ofO.1 M NaOH is titrated with 0.1 MHCI. Calculate pH when i) 20mL.
ii) 24 mL of acid is added

A

`12.0, 11.30`

B

`11.30, 12`

C

`2.0, 2.70`

D

` 2.70, 2.0`

Text Solution

Verified by Experts

The correct Answer is:
B
m mole of `NaOH = 25 xx 0.1 = 2 5`
mmol of `HCl = 20 xx 0.1 = 2`
mmol of NaOH left `= 2.5 -2.0 = 0.5` mmol
Volume `= 25 + 20 = 45 mL, [overset(o+)(O)H]= (0.5m mol)/(45mL) ~~ 0.01 = 1 ^(-2) M, pOH = 2, pH = 12 `
ii) `24 mL of 0.1 MHCl` is added.
mmol of `HCl = 24 xx 01 = 2.4,` mmol of NaOH left `= 2.5 - 2. 4 = 0.1 `
`[overset(o+)(O)H] = (0.1) /((25 + 24)) = 0.002 = 2 xx 10 ^(-3) M, pOH = - log [2 xx 10 ^(-3)] =- 0.3 + 3 = 2.7`
`pH = 14 - 2. 7 = 11.3`
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