Calculate the pH of a water at 25°C that contains 0.658 mg/L of carbonic acid (H

Question

Calculate the pH of a water at 25°C that contains 0.658 mg/L of carbonic acid (H2CO3). Assume that [H+] = [HCO3 ]. Follow the steps to solve the problem:

a.Before you attempt to solve this problem, go over example 5-8 from the textbook.

b. Consider the following reaction: H2CO3(H+)+ HCO3

c.Convert 0.658 mg/L to moles/L

d. Obtain pk a for carbonic acid from table 5-3 from the textbook

e. Transform pk a to Ka using equation 5-34: pKa= -log (Ka)

f.Do the corresponding equilibrium relationship (equation 5-33)

g.Solve for [H+] assuming that [H+] = [HCO3

h.Obtain the pH by using equation 5-29: pH = -log[H+] (answer: pH = 5.66)

Explanation / Answer

b) Reaction:

H2CO3 ---> H+ + HCO3-

c) Converting to find concentration of Carbonic Acid:

Molecular Weight of H2CO3: 62.026 g

( 0.6580 mg/L ) * ( 1 g / 1000 mg ) * ( 1 mol / 62.026 g)

[H2CO3] = 1.061 x 10^-5 mol/L

d) pKa from tables for Carbonic Acid:

pKa = 6.53

e) Using the relationship between pKa and Ka:

Ka = 10^-(pKa) <----> pka = -log (Ka)

Therefore,

Ka = 10^-6.53 = 2.95 x 10^-7

f) Using the relationship for Ka:

Ka = ([H+]*[HCO3-]) / [H2CO3]

g) Since [H+] = [HCO3], set these two concentrations as " x " in the above equation as they are unknown concentrations; replace the known concentration for the Carbonic Acid as determined above, the Ka is known as well after Step 4

Ka = (x * x) / (1.061 x 10^-5) = 2.95 x 10^-7

= x^2 = 3.13 x 10^-12

x = 1.77 x 10^-6

h) Knowing the value for x, we now now the concentration of H+; using the relationship for pH:

pH = -log [H+]

pH = -log ( 1.77 x 10^-6)

Therefore:

pH = 5.75 Ans.

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