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You have at your lab bench the following chemicals: NaH2PO4(s), Na2HPO4(s), Na3PO4(s) and deionized water. You also have standard glassware
Question
You have at your lab bench the following chemicals: NaH2PO4(s), Na2HPO4(s), Na3PO4(s) and deionized water. You also have standard glassware available. Describe how you would make 1.00 L of buffer with a pH of 7.00 using only solid materials and deionized water so that the concentration of the weak acid in the buffer is 0.150 M.
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Chemistry
4 years
2021-08-12T20:08:19+00:00
2021-08-12T20:08:19+00:00 1 Answers
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Answer:
Weigh 17.997g Na2HPO4 and 13.486g of NaH2PO4 dissolving those salts in 1.00L of deionized water
Explanation:
A buffer is made with the mixture of a weak acid and its conjugate base. We have as first to find the correct mixture based on the pKa’s of the acids:
pKa H3PO4- NaH2PO4 = 2.148
pKa NaH2PO4 – Na2HPO4 = 7.198
pKa Na2HPO4 – Na3PO4 = 12.375
A buffer works in a range of pH’s of pKa±1. Thus, we choice NaH2PO4 – Na2HPO4.
The first equation that we can write is:
NaH2PO4 = 0.150 moles –Because the concentration of the weak acid is 0.150M, in 1L = 0.150moles
Using H-H equation for this mixture:
pH = pKa + log [Na2HPO4] / [NaH2PO4]
pH is desire pH = 7.00
pKa is 7.198
[Na2HPO4] and [NaH2PO4] could be taken as the moles of each salt.
Replacing:
7.00 = 7.198 + log [Na2HPO4] / [0.150moles]
-0.198 = log [Na2HPO4] / [0.150moles]
0.6339 = [Na2HPO4] / [0.150moles]
0.095 moles = [Na2HPO4]
The mass of each salt that we must weigh is:
NaH2PO4 -Molar mass: 119.98g/mol-
0.150mol * (119.98g/mol) = 17.997g Na2HPO4
Na2HPO4 -Molar mass: 141.96g/mol-
0.095 moles * (141.96g / mol) = 13.486g of NaH2PO4
Thus, to prepare the buffer we have to:
Weigh 17.997g Na2HPO4 and 13.486g of NaH2PO4 dissolving those salts in 1.00L of deionized water