The ocean has a natural carbonate buffer system, which helps to regulate the pH

Question

The ocean has a natural carbonate buffer system, which helps to regulate the pH of seawater. a) Write the two successive dissociation reactions for carbonic acid (H2CO3), and their associated acid dissociation constants (Ka expressions). b) What is the pH of a buffer that is created by combining 65.0 ml of 1.00 M H2CO3 and 5.0 ml of 0.800 M NaHCO3? (assume the second dissociation is negligible and dont forget to account for dilution). c)What would the pH of the solution be after 1.5 ml of 2.5 M HCl is added?

Explanation / Answer

a) The dissociation reactions for carbonic acid will be:

H2CO3 --> H+ + HCO3-

Ka1 = [HCO3-][ H+] / [H2CO3]

HCO3-   ---> H+ + CO3-2

Ka2 = [CO3-2][ H+] / [HCO3-]

b) The buffer is preapred from carbonic acid and its salt (weak acid and its salt)

for buffer we know that pH is

pH = pKa + log[salt] / [acid]

Ka = 4.2 x 10-7

pKa = -logKa = 6.38

[Acid] = moles of acid / total volume = 1 X 65 millimoles / 65+5 = 65 / 70 = 0.93

[salt] = moles of salt / total volume = 5 X 0.8 / 70 = 4/70 = 0.057

pH = 6.38 + log [0.057 / 0.93] = 6.38 + log [0.057 / 0.93] = 6.38 - 1.21 = 5.17

c) if 1.5 mL of 2.5 M of HCL added thne it means

The moles of acid added = 2.5 X 1.5 millimoles = 3.75 millimoles

The Hcl will react with the salt to give acid

HCl + NaHCO3 ---> H2CO3 + Na+ + Cl-

so, the increase in moles of acid = 3.75 millimoles

final moles of acid = 65 + 3.75 = 68.75

Concentration = millimoles / total volume = 68.75 / 71.5 = 0.961

decrease in moles of salt = 4 -3.75 = 0.25

concentration of salt = 0.25 / 71.5 = 0.0035

pH = pka + log [salt] / [acid]

pH = 6.38 + log [0.0035 / 0.961] = 6.38 - 2.44 = 3.94

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