Monday, June 6, 2011

A beaker with 150 mL of an acetic acid buffer with a pH of 5.00 is sitting on ...?

A beaker with 150 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 4.10 mL of a 0.480 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760A beaker with 150 mL of an acetic acid buffer with a pH of 5.00 is sitting on ...?
let x = concentration acid

let y = concentration conjugate base



x + y = 0.100

5.00 = 4.760 + log y/x



we must solve this system



5.00 - 4.760 = log y/x

0.24 = log y/x

10^0.24 =1.74 = y/x



1.74 x = y



x + 1.74 x = 0.100

2.74 x = 0.100

x =0.0365 M = concentration acid

0.100 - 0.0365 =0.0635 M= concentration conjugate base



moles acid = 0.150 L x 0.0365 M= 0.00548

moles conjugate base = 0.0635 M x 0.150 L=0.00953



moles HCl = 4.10 x 10^-3 L x 0.480 M=0.00197



A- + H+ = HA



moles conjugate base = 0.00953 - 0.00197=0.00756

moles acid = 0.00548 + 0.00197=0.00745



total volume = 150 + 4.10 = 154.1 mL = 0.1541 L



concentration acid = 0.00745/ 0.1541 =0.0483 M

concentration conjugate base = 0.00756/ 0.1541 =0.0491 M



pH = 4.760 + log 0.0491/ 0.0483=4.77



change pH = 5.00 - 4.77=0.23
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