Calculate the ph of the solution upon the addition of 0.015 mol of naoh to the original buffer. express the ph to two decimal places.

You missed a lot of details in your question, so when we have the complete question as the attached picture so, the answer would be:

when we have the value of Ka = 1.8 x 10^-5 so, we can use it to get the Pka value

by using this formula:

Pka = -㏒Ka 

       = -㏒(1.8 x 10^-5)

       = 4.7

now, after we have got the Pka we need now to get moles of NaC2H3O2 and

moles of HC2H3O2:

when moles of NaOH = 0.015 moles

when moles NaC2H3O2 after adding NaOH
                                           
                                         = initial mol NaC2H3O2 + mol NaOH
  ∴moles NaC2H3O2      = 0.1 + 0.015 = 0.115 moles

and moles HC2H3O2 after adding NaOH

                                         = initial mol HC2H3O2 - mol NaOH

∴ moles HC2H3O2        = 0.1 - 0.015 = 0.085 moles

so, when we have moles [HC2H3O2] &[NaC2H3O2] so we can substitution its values in [A] &[HA] :

by using H-H equation we can get the PH:

when PH = Pka + ㏒[A]/[HA] 

PH = 4.7 + ㏒0.115/0.085

      = 4.8

The answer is letter c