Question

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

In the titration of 30.0 mL of a 0.200 M solution of a hypothetical compound NaH2M, what is the pH of the solution after the 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. Please help me out !

Expert Solution

[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

queen_honey_blossom answered 1 year ago

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