universalcj
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The mixture formed when
25 cm3 of 2 mol dm–3 NaOH solution ( pH : 14.3 )
is added to
50 cm3 of 2 mol dm–3 CH3COOH, for which Ka= 1.7 × 10–5 mol dm–3
( pH : 2.2 )

what is the ph of this mixture ?

the answer is 4.8

could yuo plz tell me why?
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charco
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(Original post by universalcj)
The mixture formed when
25 cm3 of 2 mol dm–3 NaOH solution ( pH : 14.3 )
is added to
50 cm3 of 2 mol dm–3 CH3COOH, for which Ka= 1.7 × 10–5 mol dm–3
( pH : 2.2 )

what is the ph of this mixture ?

the answer is 4.8

could yuo plz tell me why?
There is a reaction between the base and the acid forming a solution that contains the salt and the acid. This is called a buffer.

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universalcj
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but as you can see, the question used different volume of each solution... would it affect the result?dd
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shengoc
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(Original post by universalcj)
but as you can see, the question used different volume of each solution... would it affect the result?dd
total volume of solution is still the same for both, as all are in the same solution mixture, Vtot cancels out
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charco
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(Original post by universalcj)
but as you can see, the question used different volume of each solution... would it affect the result?dd
The total volume is required to work out the final concentrations.

1. Identify limiting reagent (it has to be the base to make an acidic buffer)
2. Determine the moles of each component remaining and formed after reaction
3. Apply ka = [H+][A-]/[HA] (rearrange to make hydrogen ion concentration the subject)
4. pH = -log10[H+]
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