Question

A beaker with 1.90×10² mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 9.00 mL of a 0.490 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Express your answer numerically to two decimal places. Use a minus ( − ) sign if the pH has decreased.

Answer 1

Ans.
let x = concentration acid
let y = concentration conjugate base

x + y = 0.100
5.00 = 4.740 + log y/x

we must solve this system

5.00 - 4.740 = log y/x
0.26 = log y/x
10^0.26 =1.82 = y/x

1.82 x = y

x + 1.82x = 0.100
2.82 x = 0.100
x =0.03546 M = concentration acid
0.100 - 0.03546 = 0.06454 M= concentration conjugate base

Volume = 1.90×102 mL = 193.8 mL = 0.1938 L

moles acid = 0.1938 L x 0.03546 M= 0.006872148
moles conjugate base = 0.06454 M x 0.1938 L =0.012507852

moles HCl = 9x 10^-3 L x .490M=0.00441

A- + H+ = HA

moles conjugate base =0.012507852 - 0.00441=0.008097852
moles acid =0.006872148+ 0.003311 =0.012507852

total volume = 193.80 + 9 = 202.8mL =0.2028 L

concentration acid = 0.012507852/ 0.2028 =0.06168 M
concentration conjugate base =0.008097852/ 0.2028 =0.0399 M

pH = 4.740 + log 0.0399   / 0.0505 = 4.637

change pH = 5.00 -4.637= 0.363
Hence , the pH will change by 0.363.