Given that Ka1 = 5.9 x 10-3 and Ka2 = 6.0 x 10-6, calculate the pH after titrating 70 mL of 0.10 M H2SO4 with 50 mL of 0.10 M KOH.
To find the pH after titrating the solutions, we need to determine the concentration of the resulting solution. Here's a step-by-step guide:
Step 1: Calculate the number of moles of H2SO4 and KOH:
Number of moles of H2SO4 = volume (L) x concentration (M)
Number of moles of H2SO4 = 0.070 L x 0.10 M
Number of moles of H2SO4 = 0.007 moles
Number of moles of KOH = volume (L) x concentration (M)
Number of moles of KOH = 0.050 L x 0.10 M
Number of moles of KOH = 0.005 moles
Step 2: Determine the limiting reactant:
Since the stoichiometric ratio between H2SO4 and KOH is 1:2, we compare the number of moles of H2SO4 to KOH:
H2SO4 : KOH = 0.007 moles : 0.005 moles
Since the number of moles of H2SO4 is greater, H2SO4 is the limiting reactant.
Step 3: Calculate the remaining moles of H2SO4:
Since all the H2SO4 reacts with KOH in a 1:2 ratio, the remaining moles of H2SO4 will be:
Remaining moles of H2SO4 = Initial moles - moles used in the reaction
Remaining moles of H2SO4 = 0.007 moles - (2 x 0.005 moles)
Remaining moles of H2SO4 = 0.007 moles - 0.010 moles
Remaining moles of H2SO4 = -0.003 moles
Note: The negative value indicates that all the H2SO4 has reacted.
Step 4: Calculate the moles of OH- ions formed:
Since KOH is a strong base, it fully dissociates, producing an equal number of OH- ions. Therefore, the moles of OH- ions formed are the same as the number of moles of KOH used:
Moles of OH- ions = 0.005 moles
Step 5: Calculate the concentration of OH- ions:
The total volume of the solution after titration is the sum of the volumes of H2SO4 and KOH:
Total volume = volume of H2SO4 + volume of KOH
Total volume = 0.070 L + 0.050 L
Total volume = 0.120 L
Concentration of OH- ions = moles of OH- ions / total volume
Concentration of OH- ions = 0.005 moles / 0.120 L
Concentration of OH- ions ≈ 0.042 M
Step 6: Calculate the pOH:
pOH = -log10[OH-]
pOH = -log10(0.042)
pOH ≈ 1.38
Step 7: Calculate the pH:
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 1.38
pH ≈ 12.62
Therefore, the pH after titrating 70 mL of 0.10 M H2SO4 with 50 mL of 0.10 M KOH is approximately 12.62.
To calculate the pH after the titration, we need to determine the concentration of the resulting solution.
Step 1: Determine the moles of H2SO4 and KOH used in the reaction.
Moles of H2SO4 = concentration of H2SO4 * volume of H2SO4 = 0.10 M * 70 mL = 0.007 moles
Moles of KOH = concentration of KOH * volume of KOH = 0.10 M * 50 mL = 0.005 moles
Step 2: Determine the limiting reactant.
Since we have a 1:1 stoichiometric ratio between H2SO4 and KOH, the reactant with fewer moles is the limiting reactant. In this case, KOH is the limiting reactant because it has fewer moles (0.005 moles).
Step 3: Determine the excess reactant.
To determine the remaining moles of the excess reactant, we subtract the moles of the limiting reactant from the initial moles of the excess reactant.
Excess moles of H2SO4 = Initial moles of H2SO4 - Moles of limiting reactant = 0.007 moles - 0.005 moles = 0.002 moles
Step 4: Determine the concentration of the excess reactant in the final solution.
The volume of the final solution is the sum of the volumes of H2SO4 and KOH used in the reaction.
Volume of final solution = volume of H2SO4 + volume of KOH = 70 mL + 50 mL = 120 mL = 0.12 L
The concentration of the excess H2SO4 in the final solution is:
Concentration of excess H2SO4 = Excess moles of H2SO4 / volume of final solution = 0.002 moles / 0.12 L = 0.0167 M
Step 5: Determine the dissociation of the excess H2SO4.
H2SO4 is a strong acid, so it will fully dissociate in water. This means that all the excess H2SO4 will yield H+ ions.
Step 6: Determine the concentration of H+ ions in the final solution.
Concentration of H+ ions = Concentration of excess H2SO4 = 0.0167 M
Step 7: Calculate the pOH of the final solution.
pOH = -log10[OH-]
Since H2SO4 is a strong acid, we can assume that all the OH- ions come from the dissociation of water.
Concentration of OH- ions = Kw / Concentration of H+ ions = 1.0 x 10^-14 / 0.0167 M = 5.988 x 10^-13 M
pOH = -log10[5.998 x 10^-13] = 12.22
Step 8: Calculate the pH of the final solution.
pH + pOH = 14
pH = 14 - pOH = 14 - 12.22 = 1.78
Therefore, the pH after titrating 70 mL of 0.10 M H2SO4 with 50 mL of 0.10 M KOH is approximately 1.78.