A beaker with 100 mL 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 0.100 M. A student adds 6.40 mL of a 0.400 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760. Express your answer numerically to two decimal places. Use a minus (-) sign if the pH has decreased.



Bonus points for this one! Thanks!!A beaker with 100 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity?
5.00 - 4.760 =0.24

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



1.74 [acetic acid] = [acetate]

[acetate]+ [acetic acid]= 0.100

1.74 [acetic acid] + [acetic acid] = 0.100

[acetic acid] = 0.0365 M

[acetate]= 0.100 - 0.0440=0.0635 M



moles acetic acid = 0.0365 M x 0.100 L = 0.00365

moles acetate = 0.0635 x 0.100 L = 0.00635



moles H+ added = 6.40 x 10^-3 L x 0.400 M=0.00256



CH3COO- + H+ = CH3COOH

moles acetate = 0.00635 - 0.00256 =0.00379

moles acetic acid = 0.00365 + 0.00256=0.00621



pH = 4.760 + log 0.00379/ 0.00621 = 4.55