Question

State with explanation, the pH of a mixture of 25.0 mL 0.50 M HOAc with 0.25 M NaOH at the endpoint. Please show work!!!

Answer 1

The reaction is

HOAc + NaOH -------------------------------> NaOAc + H2O

25x0.50 Vx0.25 0 0 initial mmoles

At the end point  

25x0.50 = Vx0.25

Thus V = 50 mL

So at the end point

HOAc + NaOH -------------------------------> NaOAc + H2O

25x0.50 Vx0.25 0 0 initial mmoles

0 0 12.5 - mmoles

thus concentration of NaOAc at the end point = mmoles/ total volume

= 12.5/(25+50)

= 0.1667 M

Now the salt is a salt of weak acid with strong base whose pH >7 due to hydrolysis and is calculated using

pH = 1/2[pkw +pka +log C]

= 1/2[14 + 4.75 + log 0.1667]

= 8.986