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How many milliliters of 4.7 M HCl must be added to 3.9 L of 0.11 M...

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Chemistry

How many milliliters of 4.7 M HCl must be added to 3.9L of 0.11 M K2HPO4 to prepare a pH =7.44 buffer?

Answer & Explanation Solved by verified expert
3.6 Ratings (552 Votes)

K2HPO4 MOLES= 3.9 x0.11= 0.429

addition HCl to K2HPO4 results in a buffer solution when HCl is added in lesser amount than K2HPO4

K2HPO4 + HCl ----------------------> KH2PO4 + KCl

0.429          a                                     0            0

0.429-a        0                                   a              a

thus buffer solution contain 'a' moles of KH2PO4 and (0.429-a) moles K2HPO4

pKa = -log Ka = -log(6.2 x10^-8) = 7.20

For acidic buffer

Henderson-Hasselbalch equation

pH = pKa + log[salt/acid]

7.44 = 7.20 +log[salt/acid]

0.24 = log[(0.429-a)/a]

[(0.429-a)/a] = 10^0.24= 1.737

0.429 -a = 1.737a

0.429= 2.737a

a= 0.157 mole

M x volume = 0.157

4.7 x V= 0.157

V= 0.0334 L

Volume = 33.4ml of HCl must be added
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