Question

A buffer solution was prepared by dissolving 4.00 grams of sodium propanate (NaCH3CH2CO2, FM 96.06 g/mol) in a solution containing 0.100 moles of propanoic acid (CH3CH2CO2H) and diluting the mixture to 1.00 L. pKa for propanoic acid is 4.874.

(a) To this solution was added 5.00mL of 1.00 M HCl. Calculate the pH of the resulting solution.

(b) To this solution was added 5.00mL of 1.00 M NaOH. Calculate the pH of the resulting solution.

Answer 1

moles of NaCH3CH2CO2 =4/96.06 = 0.041 moles

moles of propanoic acid = 0.1 moles

Voluem of mixture = 1L

So the concentration for both = [NaCH3CH2CO2] = 0.041 M and [Propanoic Acid] = 0.1 M

pKa given = 4.874

Now we add 5 ml of 1M HCL which means that there will be H+ as well as Cl- in the ions state

Which is going to convert NaCH3CH2CO2 to propanoic acid

so the concentration of salt will decrease now

Moles of Salt now =(0.0461 - 0.005 ) = 0.0366 moles

Moles of Acid now = 0.1 + 0.005 = 0.105 moles

Now the volume = 1L + 0.005 L = 1.0005

so now the pH value can be calculated by

pH = pKa + log [salt / acid ]

= pH = 4.874 + log [0.036/0.105]

= pH = 4.409

2nd case

Now as we are adding NaOH the conversion will be in the reverse direction

Propanoic acid will get converted to the salt

moles of salt = 0.041+ 0.005 = 0.046 moles

Moles of Acid now = 0.1 - 0.005 = 0.095 moles

Now the volume = 1L + 0.005 L = 1.0005

so now the pH value can be calculated by

pH = pKa + log [salt / acid ]

= pH = 4.874 + log [0.046/0.095]

=pH = 4.559