Answer to Question #286700 in General Chemistry for June

Question #286700

A 50.0 mL solution of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH.

What is the pH after 20.0 mL of base has been added? Ka CH3COOH = 1.8 x 10-5

1
Expert's answer
2022-01-12T21:30:02-0500



HA+H2OA+H3O+HA+H_2O\rightleftharpoons\>A^-+H_3O^+


Initial moles of HA (CH3_3 COOH)


=0.150×501000=0.0075moles=0.150×\frac{50}{1000}=0.0075\>moles


Moles of NAOH added

=0.150×201000=0.003moles=0.150×\frac{20}{1000}=0.003\>moles



Moles of HA remaining =0.00750.003=0.0075-0.003

=0.0045moles=0.0045\>moles



Ka=[Haq+][Aaq][HAaq]=\frac{[H^+_{aq}][A^-_{aq}]}{[HA_{aq}]}


1.8×105=[Haq+]×0.0030.00451.8×10^{-5}=\frac{[H^+_{aq}]×0.003}{0.0045}


[Haq+]=2.7×105[H^+_{aq}]=2.7×10^{-5}


PH=log2.7×105=-log\>2.7×10^{-5}

=4.5686=4.5686









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