In the titration of 30.0 mL of a 0.200 M solution of a hypothetical compound NaH2M,...

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In the titration of 30.0 mL of a 0.200 M solution of ahypothetical compound NaH2M, what is the pH of the solution afterthe addition of 30.0 mL of 0.100 M NaOH? For H3M, pKa1 = 3.00, pKa2= 6.00, and pKa3 = 9.00.?

The answer is 6.0 which is not what I got in my work. Pleasehelp me out !

Answer & Explanation Solved by verified expert
3.9 Ratings (552 Votes)

[NaH2M] = molarity x volume in Litres = 0.2 M X 0.03 L = 0.006 mol

[NaOH] = molarity x volume in Litres = 0.1 M X 0.03 L = 0.003 mol

NaH2M           + NaOH ---------------> Na2HM + H2O                   pKa2 = 6

0.006 mol            0.003 mol                          0

----------------------------------------------------------------------------

0.006 - 0.003           0                            0.003 mol

= 0.003 mol

Hence,

[NaH2M ] = 0.003 mol

[Na2HM] = 0.003 mol

pKa = pKa2 = 6

From Henderson-Hasselbalch equation,

pH = pKa + log [ conjugate base]/ [acid]

     = pKa + log [Na2HM] / [NaH2M]

      = 6 + log ( 0.003/0.003)

      = 6 + 0

      = 6

Therefore,

pH = 6


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