You have at your lab bench the following chemicals: NaH2PO4 (s), Na2HPO4 (s), Na

Question

You have at your lab bench the following chemicals: NaH2PO4 (s), Na2HPO4 (s), Na3PO4 (s), NH4Cl (s), NaCH3COO (s), CH3COOH (l), 1.00 M NH3, and deionized water. You also have standard glassware available.

H3PO4: 2.15 (pKa)

H2PO4-: 7.20 (pKa)

HPO4^2-: 12.3 (pKa)

NH4+: 9.24 (pKa)

CH3COOH: 4.75 (pKa)

a. How would you make 250.0 mL of buffer with a pH of 7.50 using only the materials listed above and deionized water so that the concentration of the acid in the buffer is 0.100M. Give the amounts of the pure materials in grams if they are solid or mL if they are liquid.

b. How would make 1.00 L of a buffer with a pH of 9.10 using only the materials listed above and deionized water so that the concentration of the base in the buffer is 0.500 M. Give the amounts of the pure materials in grams in they are solid or mL if they are liquid.

Please do not skip any steps, show all work .

Explanation / Answer

a. 250 ml of 0.1 M buffer of pH 7.50

pH = pKa + log(base/acid)

7.50 = 7.20 + log(HPO4^3-/H2PO4-)

(HPO4^2-) = 2(H2PO4-)

(HPO4^2-) + (H2PO4-) = 0.1 x 0.250 = 0.025 mol

2(H2PO4-) + (H2PO4-) = 0.025

(H2PO4-) = 0.0083 mol

mass NaH2PO4 = 0.0083 x 119.98 = 1.0 g

(HPO4^2-) = 0.025 - 0.0083 = 0.0167 mol

mass Na2HPO4 = 0.0167 x 141.96 = 2.37 g

b. 1.0 L of 0.5 M buffer of pH 9.10

pH = pKa + log(base/acid)

9.10 = 9.24 + log(NH3/NH4+)

(NH3) = 0.72(NH4+)

(NH3) + (NH4+) = 0.5 x 1 = 0.5 mol

0.72(NH4+) + (NH4+) = 0.5

(NH4+) = 0.3 mol

mass NH4Cl = 0.3 x 53.49 = 16.05 g

(NH3) = 0.5 - 0.3 = 0.2 mol

mass NH3 = 0.3/1 = 0.3 L

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