Question

A buffer solution is prepared by mixing 15.0 mL of 2.00 M Acetic Acid and 10.0 mL of 1.50 M NaC2H3O2. Determine the pH of the solution after the addition of 0.015 moles HCl (assume there is no change in volume when the HCl is added).

Ka HC2H3O2 = 1.80E-5

Answer 1

15.0 mL of 2.00 M Acetic Acid = [(15*2)/1000] =0.03 mols

10.0 mL of 1.50 M NaC2H3O2 =[(10*1.5)/1000] = 0.015 moles

After the addition of 0.015 moles HCl with the buffer solution the 0.015 moles of NaC2H3O2 will transfer into Acetic Acid.

NaC2H3O2 + HCl ----------> Acetic Acid + NaCl

After the addition of HCl , the amount of salt = (0.015-0.015)= 0 moles

After the addition of HCl , the amount of Acetic acid =(0.03+0.015) =0.045 moles

After the addition of HCl , the total amount of solution is =(10+15) =25 mL

Therefore , [acid] = (0.045*1000/25) = 1.8 M

From the dissociation of weak acid equation ,

Ka = 2C and [H+] =αC

So,

  

and   [H+] =

So,PH= -log[1.8*{10^(-5)}^(1/2)] =2.245

Therefore , the pH of the solution after the addition of 0.015 moles HCl is 2.245