You need to make 45 mM potassium phosphate buffer at pH 7.50. In your lab you se

Question

You need to make 45 mM potassium phosphate buffer at pH 7.50. In your lab you see chemical containers labeled "potassium phosphate, monobasic" (KH_2PO_4) and "potassium phosphate, dibasic" (K_2HPO_4). How many moles of each should you dissolve to make 1.0 L of buffer at the correct pH? (use the book's pKa values for phosphate: 2.14, 6.86, and 12.4) pH=pKa + log (A^-/HA) 7.50=6.86+log(A^-/HA) 7.50-6.86=log(A^-/HA) 0.64-log(A^-/HA) 10^0.64=(A^-/HA) 4.37=(A^-/HA) 4.37(HA)=(A^-/HA)(HA) 4.37HA=A^- 45mM*0.001=0.045M 4.37HA=0.045M HA=0.0084M 0.045M-0.0084M=0.0366=|0.037M KH_2PO_4=0.0084M KH_2PO_4=0.037M What is the pH of a buffer that has 0.140 moles HF (pK_a = 3.15) and 0.071 moles KF in 1.00 L of solution when 0.020 moles of HCI is added?

Explanation / Answer

we know that

for buffers

pH = pKa + log [conjugate base / acid]

in this case

acid is KH2P04

conjugate base is K2HP04

also

H2P04- ----> H+ + HP042- pKa = 6.86

given

pH = 7.5

so

7.5 = 6.86 + log [K2HP04 / KH2P04]

log [K2HP04 / KH2P04] = 7.5 - 6.86

log [K2HP04 / KH2P04] = 0.64

log [K2HP04 / KH2P04] = 10^0.64

[K2HP04 / KH2P04] = 4.36516

[K2HP04] = 4.36516 [KH2P04]

now

given that

total phosphate concentration = 45 x 10-3 M

here

total phosphate concentration = [KH2P04] + [K2HP04]

so

[KH2P04] + [K2HP04] = 45 x 10-3

[KH2P04] + 4.36516 [KH2P04] = 45 x 10-3

5.36516 [KH2P04] = 45 x 10-3

[KH2P04] = 8.39 x 10-3 M

[K2HP04] = 4.36516 x 8.39 x 10-3

[K2HP04] = 0.0366 M

now

we know that

moles = concentration x volume (L)

given

volume = 1 L

so

moles of KH2P04 = 8.39 x 10-3 x 1 = 8.39 x 10-3

moles of K2HP04 = 0.0366 x 1 = 0.0366

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