In a weak Acid-Strong Base titration, the equivalence point occurs when 12.75 mL

Question

In a weak Acid-Strong Base titration, the equivalence point occurs when 12.75 mL of 0.075M NaOH is added to a solution containing 10mL weak unknown acid and 40mL Deionized water.  

According to my graph from my experiment I think the half equivalence point is 6.375 mL with a pH of about 5.14 so pKA=5.14 ? Not sure if my calculations are correct but I did (12.75/1000) x 0.075M to get moles of NaOH and therefore moles of unknown Acid. Then divided the moles by (10mL acid/1000). Any help with explanation is appreciated, Thank you!  

Explanation / Answer

at equivalence point moles of acid = moles of base ( assuming acid is mono protic)

hence moles of acid = ( 12.75 /1000) x 0.075 = 0.00095625

vol of acid = 10 ml = 0.01 liters

Molarity of acid initially = moles/vol = 0.00095625/0.01 = 0.095625 M

but since water is added by 40 ml ,

new Molarity of diluted acid = ( 0.00095625/(0.01+0.04)) =0.019125 M

yes at half NaOH sol i.e 12.75/2 =6.375 ml in graph we have get pH

pH = pka + log [Conjugate base]/[acid]

pH = pka     , at half equivalence point where [conjugate base]=[acid] ,

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