Answer :
A 50. 0 ml sample of a 1.00 M solution of a diprotic acid is titrated with 2 M NaOH. The minimum volume of 2.00 m NaOH needed to reach a pH of 10 is 37.5 mL.
The molarity = 1 M
The volume = 50 mL
The moles of the diprotic acid = molarity × volume in L
The moles of the diprotic acid = 1 × 0.050
The moles of the diprotic acid = 0.050 mol
At the equivalence point, the moles of the diprotic acid is equals to the moles of the base = 0.050 mol.
At the 1st half equivalence:
pH = pka1 = 6
At the 2nd half equivalence :
pH = pka2 = 10
When the half reaction will occurs when we add 0.025moles of NaOH :
The total moles of NaOH = 0.075 moles
The volume of NaOH = 0.075 / 2
= 0.0375 L
= 37.5 mL
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