Explain how you would make 1.00L of 0.10M Phosphate buffer with pH 7.00. Chemica

Question

Explain how you would make 1.00L of 0.10M Phosphate buffer with pH 7.00.

Chemical Inventory: 1.0 M stock H3PO4 (aq)

                                NaH2PO4-H2O (137.99 g/mol)

                                         NaH2PO4-7H2O (268.07 g/mol)

                                 NaH3PO4-12H2O (380.12 g/mol)

I'm having a hard even beginning this one but I know I'll end up using the Henderson-Hasselbach equation at some point. Help will be greatly appreciated!

Explanation / Answer

pH = pka2 + log [ Na2HPO4] / [NaH2PO4]

since pka = 7.198 , we get pH around 7 when we use Na2HPO4 and NaH2PO4

7 = 7.198 + log [Na2HPO4.7H2O]/[NaH2PO4]

[Na2HPO4.7H2O]= 0.634 [NaH2PO4.H2O]

Na2HPO4.7H2O moles = 0.634 x NaH2PO4.H2O moles

we have 1 L of 0.1 M buffer , which means buffer moles = MV = 1 x 0.1 = 0.1

hence Na2HPO4.7H2O moles + NaH2PO4.H2O moles = 0.1

solving two equations involving Na2HPO4.7H2O and NaH2PO4.H2O we get

NaH2PO4.H2O moles = 0.0612

Na2HPO4.7H2O moles = 0.634 x0.0612 = 0.0388

hence amount of NaHPO4.H2O = 0.0612 x 137.99 = 8.445 gm

Na2HPO4.7H2O amount required = 0.0388 x 268.07 = 10.4 gm

thus 10.4 gm Na2HPO4.7H2O and 8.445 gm of NaH2PO4.H2O are taken and mixed and dissolved in water to make upto 1 liter mark. Then we get pH of solution = 7 and buffer vol is 1L with conc = 0.1 M

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