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Calculate the ph of a solution of strong dibasic base with a concentrationrnof 0.400moldm-3 at 298k. You can assume that the value of kw at thisrntemperature is 1.0 x 10-14mol2dm-6.
Accepted Answer
Steps to calculate the pH of the solution:
1. Write the dissociation equilibrium equation for the dibasic base:
B(OH)2(aq) + H2O(l) BH+(aq) + OH-(aq)
2. Set up an ICE table to represent the changes in concentrations of the species at equilibrium:
B(OH)2(aq) + H2O(l) BH+(aq) + OH-(aq)
Initial: 0.400 M 0 M 0 M
Change: -x M +x M +x M
Equilibrium: (0.400 - x) M x M x M
3. The base dissociation constant, Kb, for the dibasic base is given by:
Kb = [BH+][OH-] / [B(OH)2]
4. Substitute the equilibrium concentrations into the Kb expression and solve for x:
1.0 x 10-14mol2dm-6 = x^2 / (0.400 M - x)
5. Assuming that x is small compared to 0.400 M (which is usually the case for weak bases), we can simplify the expression to:
1.0 x 10-14mol2dm-6 ≈ x^2 / 0.400 M
6. Solving for x, we get:
x = [OH-] ≈ 1.0 x 10-7 mol dm-3
7. Finally, calculate the pH of the solution using the pOH:
pOH = -log[OH-] = -log(1.0 x 10-7 mol dm-3) = 7
pH = 14 - pOH = 14 - 7 = 7
Therefore, the pH of the solution is 7. This indicates that the solution is neutral at 298K.
1. Write the dissociation equilibrium equation for the dibasic base:
B(OH)2(aq) + H2O(l) BH+(aq) + OH-(aq)
2. Set up an ICE table to represent the changes in concentrations of the species at equilibrium:
B(OH)2(aq) + H2O(l) BH+(aq) + OH-(aq)
Initial: 0.400 M 0 M 0 M
Change: -x M +x M +x M
Equilibrium: (0.400 - x) M x M x M
3. The base dissociation constant, Kb, for the dibasic base is given by:
Kb = [BH+][OH-] / [B(OH)2]
4. Substitute the equilibrium concentrations into the Kb expression and solve for x:
1.0 x 10-14mol2dm-6 = x^2 / (0.400 M - x)
5. Assuming that x is small compared to 0.400 M (which is usually the case for weak bases), we can simplify the expression to:
1.0 x 10-14mol2dm-6 ≈ x^2 / 0.400 M
6. Solving for x, we get:
x = [OH-] ≈ 1.0 x 10-7 mol dm-3
7. Finally, calculate the pH of the solution using the pOH:
pOH = -log[OH-] = -log(1.0 x 10-7 mol dm-3) = 7
pH = 14 - pOH = 14 - 7 = 7
Therefore, the pH of the solution is 7. This indicates that the solution is neutral at 298K.
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