Question

# The next three (3) problems deal with the titration of 371 mL of 0.501 M carbonic...

The next three (3) problems deal with the titration of 371 mL of 0.501 M carbonic acid (H2CO3) (Ka1 = 4.3 x 10-7, Ka2 = 5.6 x 10-11) with 1.9 M NaOH.

1)What is the pH of the solution at the 2nd equivalence point?

2)What will the pH of the solution be when 0.1282 L of 1.9 M NaOH are added to the 371 mL of 0.501 M carbonic acid?

3)
How many mL of the 1.9 M NaOH are needed to raise the pH of the carbonic acid solution to a pH of 5.889?

1 ) second equivalence point pH = 11.96

2 ) molarity of H2CO3 = 0.371 x 0.501 / (0.371 + 0.1282) = 0.372 M

molarity of NaOH = 0.1282 x 1.9 / (0.371 + 0.1282)= 0.488 M

H2CO3 + NaoH -------------------> HCO3-

0.372 0.488 0 --------------------> initial

0 0.116 0.372 ----------------> after reaction

now anothe equilibrium

HCO3- + NaOH ---------------------> CO3-2

0.372 0.116 0

0.256 0 0.116

pH = pKa2 + log [CO3-2]/ [HCO3-]

pH = 10.25 + log (0.116/0.256)

pH = 9.90

3)

moles of H2CO3 = 0.371 x 0.501 = 0.186

H2CO3 + NaOH --------------------> HCO3-

0.186 x 0

0.186-x 0 x

pH = pKa1 + log [x /0.186-x]

5.889 = 6.37 + log [x /0.186-x]

0.330 = x / 0.186 -x

0.0614 -0.330 x = x

x = 0.0462

moles of NaOH = 0.0462

molarity = moles / volume

1.9 = 0.0462 / volume

volume = 0.02432 L

volume = 24.32 mL

volume of NaoH = 24.32 mL

#### Earn Coins

Coins can be redeemed for fabulous gifts.