A) Calculate the pH of 0.215 M carbonic acid. K a1 for carbonic acid is 4.3 X 10

ID: 1031999 • Letter: A

Question

A) Calculate the pH of 0.215 M carbonic acid. Ka1 for carbonic acid is 4.3 X 10-7.
pH = 3.52

B) Now, suppose you add some solid sodium hydrogen carbonate to the carbonic acid solution in part A). What will happen to the pH?

The pH will remain the same when the sodium hydrogen carbonate is added.

You can't tell what will happen to the pH.

The pH will rise when the sodium hydrogen carbonate is added.

A buffer is formed and the pH will rise when the sodium hydrogen carbonate is added.

The pH will fall when the sodium hydrogen carbonate is added.

A buffer is formed and the pH will fall when the sodium hydrogen carbonate is added.

A buffer is formed and the pH will remain the same when the sodium hydrogen carbonate is added.

C) Calculate the pH of solution when the concentration of sodium hydrogen carbonate is 0.820 M.
pH = 6.94790693688853

Please show work. Correct answers are listed.

a) [H2CO3] = 0.215 M

H2CO3 <==> H+ + HCO3-

let x amount dissociated

Ka = [HCO3-][H+]/[H2CO3]

4.3 x 10^-7 = x^2/0.215

x = [H+] = 3.04 x 10^-4 M

pH = -log[H+] = 3.52

B) If sodium hydrogen carbonate is added to part A)

The pH will rise when sodium hydrogen carbonate is added

A buffer will form and pH will rise.

C) [HCO3-] = 0.820 M

[H2CO3] = 0.215 M

pKa = 6.37

Using Hendersen-Hasselbalck equation,

pH = pKa + log(HCO3-/H2CO3)

= 6.37 + log(0.820/0.215)

= 6.95

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