Wednesday, September 21, 2011

Adding strong acid to buffer?

A beaker with 105ml of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100M . A student adds 6.60ml of a 0.360M solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.Adding strong acid to buffer?
5.00 - 4.76=0.24



10^0.24 =1.74 = [acetate]/ [acetic acid]



[acetate]= 1.74 x [acetic acid]



[acetate] + [acetic acid]= 0.100



[acetic acid]= 0.100- [acetate]



[acetate]= 1.74 ( 0.100 - [acetate] = 0.174 - 1.74 acetate



[acetate]= 0.0635 M

[acetic acid ]= 0.0365 M



moles acetate = 0.105 L x 0.0635 =0.00667

moles acetic acid = 0.0365 x 0.105 L=0.00383



now the student add 6.60 mL of a 0.360 M of ???

hope helps
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