Question

Consider the titration of 100.0 mL of 0.75 M H₃A by 0.75 M KOH for the next three questions. The triprotic acid has K_a1 = 1.0 x 10^-5, K_a2 = 1.0 x 10^-8, and an unknown value for K_a3.

1) Calculate the pH after 100.0 mL of KOH has been added.

pH =

Tries 0/45

2) Calculate the pH after 150.0 mL of KOH has been added.

pH =

Tries 0/45

3) The pH of the solution after 200.0 mL of KOH has been added is 10.00. Determine the value of K_a3 for this triprotic acid. Use scientific notation to enter this answer, e.g., 1.0 x 10^-3 = 1.0E-3.

Ka₃ =

Answer 1

ANSWER - Question 1: Addition of 100.0 mL of KOH

Initially we have 100.0 mL of 0.75 M H₃A, this means that the moles of H₃A are:

When 100.0 mL of 0.75 M KOH are added, the moles of OH^- are:

The reaction of H₃A with OH^- is:

	H3A	+	OH-	--->	H2A-	+	H2O
Initial	0.075 mol		0.075 mol		0 mol		---
Change	-0.075 mol		-0.075 mol		+0.075 mol		---
Final	0 mol		0 mol		0.075 mol		---

At this point we have 0.075 mol of H₂A^- in 200 mL of solution (volume of KOH + volume of triprotic acid). All H₃A had been consumed. The concentration of H₂A^- is:

The pH of a solution 0.375 M of H₂A^- could be calculated using the equilibrium reaction for this especie. There are two possibilities:

The first reaction has the highest equilibrium constant. Then, the first reaction prevail over the second reaction.Therefore,

	H2A-	+	H2O	--->	HA-2	+	H3O+
Initial	0.375 M		---		0 M		0 M
Change	-X		---		+X		+X
Equilibrium	0.375 - X M		---		X M		X M

Finding the X value, we can calculate the pH. The value of X could be found with the equilibrium constant expression:

the value of Ka² is very small, this means that X will be a very small value (X << 0.375 M) and we can made an approximation: 0.375 - X ≈ 0.375. Then,

Now we can calculate the pH of solution:

After 100.0 mL of KOH has been added, the pH of solution is 4.21

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ANSWER - Question 2: Addition of 150.0 mL of KOH

From question 1, we know that the first 100 mL of 0.75 M KOH (0.075 mol of OH^-), will transform all H₃A (0.075 mol) into H₂A^-:

	H3A	+	OH-	--->	H2A-	+	H2O
Initial	0.075 mol		0.075 mol		0 mol		---
Final	0 mol		0 mol		0.075 mol		---

At this moment, it remains 50.0 mL of 0.75 M KOH. This will react with the H₂A^- into HA^-2

	H3A	+	OH-	--->	H2A-	+	H2O
Initial	0.075 mol		0.0375 mol		0 mol		---
Change	-0.0375 mol		-0.0375 mol		+0.0375 mol		---
Final	0.0375 mol		0 mol		0.0375 mol		---

At this point we have 0.0375 mol of H₂A^- and 0.0375 mol of HA^-2in 250 mL of solution (volume of KOH + volume of triprotic acid). We have both, an acid (H₂A^-) and its conjugated base (HA^-2), then we have a buffer,

the pH of a buffer could be calculated with the Hendersson-Hasselbach equation:

We have the same amount of H₂A^- (0.0375 mol) and HA^-2 (0.0375 mol) in 250 mL of solution. Then,

Therefore,

After 150.0 mL of KOH has been added, the pH of solution is 8.00

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ANSWER - Question 3: Addition of 200.0 mL of KOH

From question 1, we know that the first 100 mL of 0.75 M KOH (0.075 mol of OH^-), will transform all H₃A (0.075 mol) into H₂A^-:

	H3A	+	OH-	--->	H2A-	+	H2O
Initial	0.075 mol		0.075 mol		0 mol		---
Final	0 mol		0 mol		0.075 mol		---

and the other 100.0 mL of 0.75 M KOH (0.075 mol of OH^-), will transform all H₂A^- (0.075 mol) into HA^-2:

	H2A-	+	OH-	--->	HA-2	+	H2O
Initial	0.075 mol		0.075 mol		0 mol		---
Final	0 mol		0 mol		0.075 mol		---

At this point we have 0.075 mol of HA^-2 in 300 mL of solution (volume of KOH + volume of triprotic acid). All H₃A and H₂A^- had been consumed. The concentration of HA^-2 is:

The pH of a solution 0.25 M of HA^-2 is 10.00, this means that [H₃O⁺] is:

This is the concentration of H₃O⁺ in equilibrium. Then, at equilibrium we have:

	HA-2	+	H2O	--->	A-3	+	H3O+
Initial	0.25 M		---		0 M		0 M
Change	-1.00x10-10 M		---		+1.00x10-10 M		+1.00x10-10​​​​​​​ M
Equilibrium	0.25 M		---		1.00x10-10​​​​​​​ M		1.00x10-10​​​​​​​ M

Now, we can calculate Ka₃

The value of Ka₃ for the triprotic acid H₃A is 4x10^-20