Question

In: Chemistry

(a) Which two of the following compounds would you mix to make a buffer of pH...

(a) Which two of the following compounds would you mix to make a buffer of pH 7.45? H₃PO₄ (FM 98.00), NaH₂PO₄ (FM 119.98), Na₂HPO₄ (FM 141.96), and Na₃PO₄ (FM 163.94)?

NaH₂PO₄ and Na₂HPO₄ This is the correct answer for this part.

(b) If you wanted to prepare 1.00 L of buffer with a total phosphate concentration of 0.0513 M, how many grams of the two selected compounds would you mix together?

acidic component	_____g
basic component	______g

(c) If you did what you calculated in part (b), you would not get a pH of exactly 7.45. Explain how you would really prepare this buffer in the lab.

Solutions

Expert Solution

solution:

The pKa of given acids are

H3PO4 = 2.14

H2PO4- = 7.10

HPO4-2 = 12.32

so we will take a combination of NaH₂PO4 and Na₂HPO4, as the pKa of NaH₂PO4 is nearest to the desired pKa value.

b) The total volume required = 1L

The total concentration required = 0.0513

it means

[NaH₂PO4] + [Na₂HPO4] = 0.0513 .......(1)

From Hendersen Equations

pH = pKa + log [salt] / [acid] = 7.10 + log [Na₂HPO4] / [NaH₂PO4] = 7.45

log [Na₂HPO4] / [NaH₂PO4] = 0.35

[Na₂HPO4] / [NaH₂PO4] = 2.24 ......(2)

Putting (2) in (1)

2.24 NaH₂PO4]+[NaH₂PO4] = 0.0513

[NaH₂PO4] = 0.0154 M

[Na₂HPO4] = 0.050 - 0.0154 = 0.0346 M

as the solution is 1L

so the moles of Na₂HPO4 required = 0.0346 moles

Mass of Na₂HPO4 = moles X molar mass = 0.0346 moles X 141.96 = 4.91 grams

Moles of NaH₂PO4 = 0.0154

Mass of NaH₂PO4 = 0.0154 X 119.98 = 1.85 g

c) we will take the salt initially i.e. Na₂HPO4, 0.0513moles

It means we will weight out = 0.0513 X 141.96 = 7.28 grams of salt

And will dissolve it in 900mL of water. then we will start adding HCl to it so that HCl will react with the salt to give the weak acid. and will continue adding HCl till we get desired pH. Then we will make up water to 1L.

queen_honey_blossom answered 1 year ago

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