A beaker with 185 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity o?

A beaker with 185 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 4.9 mL of a 0.290 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.A beaker with 185 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity o?
let x = [acetic acid]

let y = [acetate]



x + y = 0.100



5.00 = 4.760 + log y/x



0.24 = log y/x

10^0.24 = y/x =1.74



y = 1.74 x



x + 1.74 x = 0.100



2.74 x = 0.100



x = 0.0365 M

y = 0.0635 M



moles acetic acid = 0.0365 x 0.185 L = 0.00675



moles acetate = 0.0635 x 0.185 L=0.0117



moles H+ added = 4.9 x 10^-3 L x 0.290 = 1.42 x 10^-3



moles acetic acid = 0.00675 + 1.42 x 10^-3 = 0.00817

moles acetate = 0.0117 - 1.42 x 10^-3=0.0103



pH = 4.760 + log 0.0103/ 0.00817= 4.86