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Question

100ml of 0.1 M NaOH is added to 100 ml of a 0.2 M CH3COOH solution. The pH of the resulting solution will be (pka=4.74)


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Solution

Millimole of 100ml of 0.1 M NaOH=100x0.1=10 millimole.

Millimole of 100 ml of a 0.2 M CH3COOH =100x 0.2=20 millimole.

NaOH(aq.)SodiumHydroxide+CH3COOH(aq.)AceticacidCH3COONa(aq.)SodiumAcetate+H2O(l)Water10millimole20millimole010millimole10millimole

Now it is a Buffer.

pH=pka+logSaltAcid

pH=4.74+log1010=4.74+log1=4.74+0=4.74

100ml of 0.1 M NaOH is added to 100 ml of a 0.2 M CH3COOH solution. The pH of the resulting solution will be 4.74.


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