Question

# Prepare 250.00 mL of an phosphate buffer (pKa=7.199) with a total phosphate ion concentration of 0.020000...

Prepare 250.00 mL of an phosphate buffer (pKa=7.199) with a total phosphate ion concentration of 0.020000 M and a pH of 7.100.

Calculate the amount of the Monobasic Potassium Phosphate (KH2PO4 FW=136.086) required to prepare this solution. Report your answer in grams.

____________ grams

Calculate the amount of Dibasic Sodium Phosphate (Na2HPO4 FW=141.959) required to prepare this solution. Report your answer in grams.

____________ grams

the pKa is given as 7.199 for:

HPO4-2 and H2PO4-

then

pH = pKa2 + log(HPO4-2/ H2PO4-)

7.10 = 7.199 +  log(HPO4-2/ H2PO4-)

(HPO4-2/ H2PO4-) = 10^(7.10-7.199)

(HPO4-2/ H2PO4-) = 0.796

we know that

(HPO4-2 + H2PO4-) = 0.02*250 = 5 mmol

from ratio

(HPO4-2/ H2PO4-) = 0.796

HPO4-2 = 0.796* H2PO4-

(HPO4-2 + H2PO4-) = 5

0.796* H2PO4- + H2PO4- = 5

H2PO4- = 5/(0.796+1) = 2.7839 mmol

KH2PO4 = 2.7839 mmol

mass = mmol*MW = 2.7839*136.086 = 378.849 mg = 0.378849 g

now..

(HPO4-2 + H2PO4-) = 5

HPO4- = 5-2.7839 = 2.2161

Na2HPO4 = 2.2161 mmol

mass = mmol*MW = 2.2161*141.959 = 314.595 mg = 0.314595 g of Na2HPO4

#### Earn Coins

Coins can be redeemed for fabulous gifts.