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In an acid base titration experiment, 50.0 ml of a 0.0500 m solution of acetic acid...

In an acid base titration experiment, 50.0 ml of a 0.0500 m solution of acetic acid ( ka =7.5 x 10^-5) was titrated with a 0.0500 M solution of NaOH at 25 C. The system will acquire this pH after addition of 20.00 mL of the titrant:

Answer is 4.581

Please show all work

Expert Solution

No. of moles of acetic acid initially;

No of moles of NaOH added;

Now, draw the ice table for the reaction,

	CH₃COOH
Initially	2.5*10^-3mol
When NaOH is added	2.510^-3-110^-3 mol
Finally	1.5*10^-3 mol

The conentration of acetic acid= 1.5/(50+20)=0.0214 M

Consider the dissociation of acetic acid;

CH3COOH	CH3COO^- +	H⁺
0.0214 M	0	0
0.0214-X	X	X

As x <<< 0.0214, we assume that 0.00214=0.00214-x.

Now, calculate the pH of the system using the fllowing equation,

queen_honey_blossom answered 3 years ago

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