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CHM 221 - Fall 17 - PONTON > Activities and Due Dates > Polyprotic Acids and Bas

ID: 547761 • Letter: C

Question

CHM 221 - Fall 17 - PONTON > Activities and Due Dates > Polyprotic Acids and Bases 10/27/2017 0800 AM 0 32/44 0 10/25/2017 11:21 AM A Print I Calculator a A Periodic Table tion 6 of 11 a Sapling learning Phosphoric acid is a triprotic acid with the following pK, values pK.2.148 pK7,198 pK, -12.375 a 2 You wish to prepare 1.000L of a 0.0500 M phosphate buffer at pH 7.600. To do this, you choose to mix the two salt forms involved in the second ionization, Nat2POs and NazHPO4in a 1.000 L volumetric flask and add water to the mark. What mass of each salt will you add to the mixture? Number Number mass NaH, PO ,-0 mass Na, HPO, - IU What other combination of phosphoric acid andor its salts could be mixed to prepare this buffer? (Check all that apply). OHPO4 and NahzPO4 HBPO4 and NazHPO. D and NauPO4 H3PO, O NaPO4 and NauPO, O NaHPO, and NauPO. Hint NO Previous O Check Answer O Next Exit

Explanation / Answer

a)

Answer

Mass of NaH2PO4 to be added = 1.7037g

Mass of Na2HPO4 to be added = 5.0822g

Explanation

Henderson - Hasselbalch equation

pH = pKa + log([A-]/[HA])

A- = HPO42-

HA = H2PO4-

pKa of H2PO4-

required pH is 7.600

Therefore,

7.6 = 7.198 + log ( [ HPO42-]/[H2PO4-] )

log([HPO42-]/[H2PO4-] ) = 0.402

[ HPO42-] / [H2PO4-] = 1×10^0.402 = 2.523

[ HPO42- ] = 2.523 [ H2PO4-]

Total concentration of Buffer = 0.0500M

Therefore,

[ H2PO4- ] + 2.523 [ H2PO4-] = 0.0500M

3.523 [ H2PO4-] = 0.0500M

[ H2PO4-] = 0.0142M = 0.0142 mol /Liter

So ,

[ HPO42- ] = 0.0500 - 0.0142M

= 0.0358M = 0.0358 mol /liter

Requred buffer volume = 1L

Therefore,

mole of H2PO4- to be added = 0.0142

mole of HPO42- to be added = 0.0358

mole of H2PO4- = mole of NaH2PO4

mole of HPO42- = mole of Na2HPO4

Mass = Molar mass × No of mole

Molar mass of NaH2PO4 = 119.98g/mol

Mass of NaH2PO4 to be added = 0.0142mol × 1198.98g/mol = 1.7037g

Molar mass of Na2HPO4 = 141.96g/mol

mass of Na2HPO4 to be added = 0.0358mol × 141.96g/mol = 5.0822g

b) H3PO4 and Na2HPO4