Consider a the titration of 0.969 L of 0.891 M carbonic acid (H2CO3) with 1.85 M...

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Answer 1

Answer #1

No of mol of H2CO3 = 0.969*0.891 = 0.863 mol

1 mol H2CO3 = 2 mol NaOH

no of mol of NaOH required to reach second equivalence point = 2*0.863 = 1.726 mol

volume of NaOH must take = 1.726/1.85 = 0.933 L

total volume of mixture = 0.969+0.933 = 1.902 L

concentration of salt(Na2CO3) = 0.863/1.902 = 0.454 M

pH of salt = 7+1/2(pka2+logC)

pka2 of H2CO3 = 10.33

pH = 7+1/2(10.33+log0.454) = 12

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Answer 2

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