A 50.0 ml sample of a 1.00 m solution of a diprotic acid h2a (ka1 = 1.0 × 10–6 and ka2 = 1.0 × 10–10) is titrated with 2.00 m naoh. what is the minimum volume of 2.00 m naoh needed to reach a ph of 10.00?

37.5mL of 2.00M NaOH

Explanation:

The reactions of the diprotic acid (H₂A) with NaOH are:

1) H₂A + NaOH → HA⁻ + H₂O + Na⁺ Ka = 1x10⁻⁶; pKa = 6

2) HA⁻ + NaOH → A²⁻ + H₂O + Na⁺ Ka = 1.0x10⁻¹⁰; pKa = 10

First, you need to add NaOH until the first reaction occurs completely, that is the addition of:

0.050L H₂A ₓ (1mol / 1L) = 0.050 mol H₂A ≡ 0.050mol NaOH

Now, in the second reaction you need the half of the reaction occurs, because in this point: HA⁻ = A²⁻ that means pH = pKa

It is a buffer when HA⁻ is the weak acid and A²⁻ is conjugate base, the theory says that when weak acid = conjugate base; pH = pKa.

As the complete reaction needs 0.050moles of NaOH, the half reaction occurs when you add 0.025moles of NaOH. Thus, total moles you need to add are:

0.075 moles of NaOH. As the base is 2.00M:

0.075 moles of NaOH × (1L / 2.00mol) = 0.0375L ≡ 37.5mL of 2.00M NaOH

the potential energy of the reactants is 200j.

from the energy diagram, the energy of the product formed is 350j; this means that, this reaction is an endothermic reaction, because it absorbs energy from its environment.