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Answer :
Answer:
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
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When The minimum volume of 2.00 m naoh needed to reach a ph of 10.00 is = 37.5mL of 2.00M NaOH
What is Diprotic acid?
When The reactions of the diprotic acid (H₂A) with NaOH are:
1) Then H₂A + NaOH → HA⁻ + H₂O + Na⁺ Ka = 1x10⁻⁶; pKa = 6
2) After that HA⁻ + NaOH → A²⁻ + H₂O + Na⁺ Ka = 1.0x10⁻¹⁰; pKa = 10
When we need to add NaOH until the first reaction occurs completely, that is the addition of:
Then 0.050L H₂A ₓ (1mol / 1L) = 0.050 mol H₂A ≡ 0.050mol NaOH
Now, in the second reaction When you need the half of the reaction occurs, because in this point: HA⁻ = A²⁻ that means pH = pKa
When It is a buffer when HA⁻ is the weak acid and A²⁻ is conjugate base, the theory says that when weak acid is = conjugate base; pH = pKa.
Although As the complete reaction needs 0.050moles of NaOH, When the half reaction occurs when you add 0.025moles of NaOH. Thus, total moles you need to add are:
Then 0.075 moles of NaOH. As the base is 2.00M:
Therefore, 0.075 moles of NaOH × (1L / 2.00mol) = 0.0375L ≡ 37.5mL of 2.00M NaOH.
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