Need parts B and C answered only. Thank you. 9-20. (a) Which two of the followin

Question

Need parts B and C answered only. Thank you.

9-20. (a) Which two of the following compounds would you mix to make a buffer of pH 7.45: H3PO (FM 98.00), NaH2PO4 (FM 119.98), Na2HPO4 (FM 141.96), and Na3PO (FM 163.94)? (b) If you wanted to prepare 1.00 L of buffer with a total phosphate concentration of 0.050 0 M, how many grams of each of the two selected compounds would you mix together? (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.

Explanation / Answer

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 = 0.050

it means

[NaH2PO4] + [Na2HPO4] = 0.050 .......(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] = 0.050

[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, 0.05 moles

It means we will weight out = 0.05 X 141.96 = 7.098 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.

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.