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25.00 mL of 0.150M benzoic acid HC7H5O2 was titrated with 0.200 M NaOH. Calculate the pH...

25.00 mL of 0.150M benzoic acid HC7H5O2 was titrated with 0.200 M NaOH. Calculate the pH at the following points. Ka for benzoic acid is 6.3x10-5

a. Before adding any NaOH

b. Halfway to equivalence

c. After adding 12.2 mL of the NaOH

d. At the equivalence point

Please answer all the parts!

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Answer #1

Question is solved using concepts of equilibrium.

Salution- The giren data are - volume of benzoic aud - 25.00 vel Comunbeatie-(Malawuty of benzola acid z o SOM concentrationHence [ut] z 2 z 7.8X10 25 me milimade 3-12x163M pH = -log.hu) 2 - log [3-12X103] 2 –109 312 +3 log18 23 - 0.4944 22:50 at pRINTTTT C - after adding 12-2nd of the Naon milimale of Naon added - 12.2nd X 0.200M - z 2.98 milimale Cook G + NaOH = CooNaI W At the equivalence point Milimale of Acid z Milimcles of sabuse 25 X 0.150 z 0.200 X V V z 25 X 6.150 0.200 V z 18-75 ml

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