A1.0l buffer solution contains 0.100 mol of hc2h3o2 and 0.100 mol of nac2h3o2. the value of ka for hc2h3o2 is 1.8×10−5. part a 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.

Hey there!:

Ka = 1.8*10⁻⁵

Pka = 4.74

add the moles of  NaOH to moles of base:

0.015 moles + 0.100 moles => 0.115 moles

Since you add mol base you have to take away that same , amount from acid:

0.100 acid  - 0.015  NaOH = 0.085

Therefore :

pH = pKa + log [ A ] / [ HA ]

pH = 4.74 + log [ 0.115 ] / [ 0.085 ]

pH =  4.87600

Hope that helps!

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