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

answer: at room temperature (roughly 20 degrees c), you can dissolve 203.9 grams (or 200 grams) of sugar in 100 ml of water. as water temperature increases, the amount of sugar that will dissolve increases, up to about 500 grams of sugar in 100 ml water at 100 degrees c.

explanation: