Question

A beaker with 2.00×10² 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 4.40 mL of a 0.390 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

Using Henderson-hasselbach equation, pH=pka+log [base]/[acid]

Or, 5.00=4.74+log[acetate]/[acetic acid]

Or,0.26= log[acetate]/[acetic acid]

[acetate]/[acetic acid]=10^0.26=1.82

or, [acetate]/[acetic acid]=1.82

As total molarity of the buffer=0.1M=0.1 mol/L

[acetate]+[acetic acid]=0.1 mol/L

Or, [acetate]=1.82[acetic acid]=1.82 (M-[acetate]=0.182-1.82[acetate]

Or, [acetate]=0.182-1.82[acetate]

2.182 [acetate]=0.182M

[acetate]=0.182M/2.182=0.0834 M

Molarity of acetic acid=0.1-0.0834=0.0166M

As the total volume of buffer=2*102ml=204ml

Moles of acetic acid=0.0166M*204ml*10^-3 L/ml=3.386 *10^-3 mol

Moles of acetate=0.0834 mol/L*204*10^-3L=17.0136 *10^-3 mol

Now, as Moles ofHCl added=4.40ml*0.390M=4.40*10^-3 L*0.390 mol/L=1.716*10^-3 moles

So moles of acetate neutralized =1.716*10^-3 moles

New moles of acetate=17.0136 *10^-3 mol-1.716*1.716*10^-3 moles

=15.298**10^-3 moles

Acetate +HCl=acetic acid +cl-

New moles of acetic acid=3.386 *10^-3 mol+1.716*10^-3 moles=5.102*10^-3 moles

Using Henderson-hasselbach equation, pH=pka+log [base]/[acid]

New pH=4.74+ log (15.298**10^-3 moles)/( 5.102*10^-3 moles

         Ph=4.74+log 2.998=4.74+0.477=5.22

pH=5.22

pH=5.00+0.22