Frequently Asked Question

Here's how to calculate the pH for each scenario:
a) After adding 10.00 mL of 0.005 M KOH
1. Calculate moles of H2CO3 and KOH:

moles H2CO3 = (0.01 M)
(0.020 L) = 2.0 x 10^-4 moles

moles KOH = (0.005 M)
(0.010 L) = 5.0 x 10^-5 moles
2. Determine the limiting reactant: KOH is the limiting reactant since it's present in fewer moles.
3. Calculate the moles of H2CO3 remaining:

moles H2CO3 remaining = 2.0 x 10^-4 moles - 5.0 x 10^-5 moles = 1.5 x 10^-4 moles
4. Calculate the concentration of H2CO3:

[H2CO3] = (1.5 x 10^-4 moles) / (0.030 L) = 5.0 x 10^-3 M
5. Use the Ka expression to find [H+] and then pH:

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

Assuming [H+] = [HCO3-], we have Ka = [H+]^2 / [H2CO3]

[H+] = sqrt(Ka
[H2CO3]) = sqrt(4.6 x 10^-7
5.0 x 10^-3) = 1.5 x 10^-5 M

pH = -log(1.5 x 10^-5) = 4.82
b) After adding 20.00 mL of 0.005 M KOH
1. Calculate moles of KOH:

moles KOH = (0.005 M)
(0.020 L) = 1.0 x 10^-4 moles
2. Calculate the moles of H2CO3 remaining:

moles H2CO3 remaining = 2.0 x 10^-4 moles - 1.0 x 10^-4 moles = 1.0 x 10^-4 moles
3. Calculate the concentration of H2CO3:

[H2CO3] = (1.0 x 10^-4 moles) / (0.040 L) = 2.5 x 10^-3 M
4. Use the Ka expression to find [H+] and then pH:

[H+] = sqrt(Ka
[H2CO3]) = sqrt(4.6 x 10^-7
2.5 x 10^-3) = 1.1 x 10^-5 M

pH = -log(1.1 x 10^-5) = 4.96
c) At the equivalence point
1. Determine the volume of KOH needed to reach the equivalence point:

moles H2CO3 = moles KOH

(0.01 M)
(0.020 L) = (0.005 M)
V

V = 0.040 L or 40.00 mL
2. Calculate the total volume at the equivalence point:

Vtotal = 20.00 mL + 40.00 mL = 60.00 mL
3. Calculate the concentration of the conjugate base (HCO3-) at the equivalence point:

[HCO3-] = (2.0 x 10^-4 moles) / (0.060 L) = 3.3 x 10^-3 M
4. Use the Kb expression for HCO3- to find [OH-] and then pH:

Kb = Kw/Ka = 1.0 x 10^-14 / 4.6 x 10^-7 = 2.2 x 10^-8

Kb = [OH-][H2CO3] / [HCO3-]

Assuming [OH-] = [H2CO3], we have Kb = [OH-]^2 / [HCO3-]

[OH-] = sqrt(Kb
[HCO3-]) = sqrt(2.2 x 10^-8
3.3 x 10^-3) = 8.5 x 10^-6 M

pOH = -log(8.5 x 10^-6) = 5.07

pH = 14 - 5.07 = 8.93
Important Note: These calculations assume that the second ionization of carbonic acid (H2CO3 → HCO3- → CO3^2-) is negligible. This is a reasonable assumption because the second Ka value is much smaller than the first Ka value. The pH values calculated here are approximate, and a more accurate calculation would consider the second ionization.