Question

a.) How would you prepare .250 L of a 0.300 M phosphate buffer at pH=3 (H3PO4: pka1 = 2.12, pka2 = 7.21, pka = 12.32) using the appropriate weak acid and conjugate salt. b.) How would you prepare the assigned buffer (given in question #1a) if using combining the weak acid and 0.400 M NaOH? c.) How would you prepare the assigned buffer (given in question #1a) if using combining the weak acid and 0.400 M HCl?

Answer 1

a) 0,857 g of H₃PO₄ with 9,016 g of KH₂PO₄

b) 166 mL of 0,400M NaOH

c) 22 mL of 0,400M HCl

Step-by-step explanation:

a) The appropriate weak acid and conjugate salt are:

H₃PO₄ ⇄ H₂PO₄⁻ + H⁺ where pka = 2,12

Henderson–Hasselbalch equation finding pH = 3:

3 = 2,12 + log₁₀
`([H2PO4-])/([H3PO4])`

7,59 =
`([H2PO4-])/([H3PO4])` (1)

If buffer concentration is 0,300M:

0,300 M = [H₃PO₄] + [H₂PO₄⁻] (2)

Replacing (2) in (1):

[H₃PO₄] = 0,035 M

Thus:

[H₂PO₄⁻] = 0,265 M

Thus, to prepare this buffer you need weight:

0,035 M × 0,250 L = 8,75x10⁻³ moles ×
`(97,994 g)/(1mol)` = 0,857 g of H₃PO₄

And:

0,265 M × 0,250 L = 6,63x10⁻² moles ×
`(136,086 g)/(1mol)` = 9,016 g of KH₂PO₄

b) Using 0,400 M NaOH the equilibrium is:

H₃PO₄ + NaOH ⇄ H₂PO₄⁻ + H₂O

Knowing the equilibrium concentrations are:

[H₃PO₄] = 0,035 M = 0,300 M - x -Because in the first all 0,300 M must be of H₃PO₄-

[H₂PO₄⁻] = 0,265 M

Thus, x = 0,265 M are NaOH needed to obtain the desire pH. Those are obtained thus:

0,265 mol/ L × 0,250L ×
`(1L)/(0,400mol)` = 0,166 L ≡ 166 mL of 0,400M NaOH

c) Using 0,400 M HCl the equilibrium is:

H₃PO₄ ⇄ H₂PO₄⁻ + HCl

Knowing the equilibrium concentrations are:

[H₃PO₄] = 0,035 M

[H₂PO₄⁻] = 0,265 M = 0,300 M - x -Because in the first all 0,300 M must be of H₂PO₄-

Thus, x = 0,035 M are HCl needed to obtain the desire pH. Those are obtained thus:

0,035 mol/ L × 0,250L ×
`(1L)/(0,400mol)` = 0,022 L ≡ 22 mL of 0,400M HCl

I hope it helps!