Question

A beaker with 1.20×10² 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 7.20 mL of a 0.330 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

The pH of the buffer is given by the following reaction:

pH = pKa + log (base/acid)
Here, pH = 5.000 and The pKa of acetic acid is 4.740.

First calcualte the the ratio of base to acid as follows:

5.00 = 4.740 + log (x/1)

log x = 0.26

x = 1.82
Therefore the base:salt ratio is 1.82:1


assume that if x = moles of acid

1.82x + x = 0.100 total moles of solute

x= 0.100/2.82=0.0355 means mole of acid

and base = 0.0646 mol

Now aad strong acid, HCl 7.20 mL of a 0.330 M this beaker.


(0.330 mol/L) (0.00720 L) = 0.002376 mol of strong acid added


new mole of acid and base will be:
base: 0.0 646- 0.002376 = 0.06224

acid: 0.0355 + 0.002376 = 0.037876


pH = 4.740 + log(0.06224/0.037876)

pH = 4.740 + log(0.06224/0.037876)

pH = 4.740 + 0.216

pH = 4.956

decreased pH = 5.000 -4.956= 0.044