How many grams of carbonic acid mus tbe added to 100 mL of 0.5M NaHCO3 (Ka1=...

Question

Question

How many grams of carbonic acid mus tbe added to 100 mL of 0.5M NaHCO3 (Ka1= 4.5 x10 -7 Ka2= 4.7 x 10-11) tp get a pH of 6.5?

science chemistry

Answer 1

Answer #1

Ka1= 4.5 x10 -7

PKa1 = - log Ka1 = - log (Ka1) = 4.5 x10 -7 ) = 6.35

100 mL of 0.5M NaHCO3 = 0.100 * 0.5 = 0.05 mole.

Resulting mixture is buffer solution.

PH = PKa1 + log [salt] / [acid]

6.5 = 6.35 + log (0.05 / [H2CO3]

[H2CO3] = 0.0354 mole H2CO3

0.0354 mole H2CO3 = 0.0354 * 62.03 = 2.2 gm

2.2 gm grams of carbonic acid mus tbe added.

Answer 2

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