1. Buffers
1986 A

2. Question
In water, hydrazoic acid, HN3, is a weak acid that has an equilibrium constant, Ka, equal to 2.8 x 10-5 at 25°C. A .300 L sample of a .050 M solution of the acid is prepared
(a) Write the expression for the equilibrium constant, Ka, for hydrazoic acid
(b) Calculate the pH of this solution at 25°C
(c) To .150 L of this solution, .80 gram of NaN3 is added. The salt dissolved completely. Calculate the pH of the remaining solution at 25°C is the volume of the solution remains unchanged.
(d) To the remaining .150 L of the original solution, .075 L of .100 M NaOH solution is added. Calculate the [OH-] for the resulting solution at 25°C

3. Part A
To write the equilibrium constant, we first write an equation describing the reaction taking place
HN3 ↔ H+ + N3-
The equilibrium constant is equal the concentration of the products raised to their exponents over the concentration of the reactants raised to their exponents.
Therefore, in this case, the Ka is equal to [H+ ][N3-]/[HN3]

4. Part B The pH of a solution is defined as –log[H+]
To solve this problem then, we need to find the concentration of H+. We can do this by creating an ICE table.
Our ICE table will be three columns by three rows. We label the first row Initial Concentration, the second row Change in Concentration, and the final row Equilibrium Concentration. We label the columns HN3, H+, and N3-, the three aqueous substances involved in our chemical equation.
The initial concentration of the HN3, according to the problem, is .05 M. The initial concentrations of the other substances is 0.
Because all the coefficients in the equation are 1, the mole to mole ratio of each substance is 1. Therefore, the changes that occur in equilibrium in the concentration of each substance are equal. The changes can be represented by the variable x. Because initially there is only HN3 present, the HN3 will decrease in concentration by x amount and both the H+ and the N3- increase in concentration by x amount.

5. Part B (continued) HN3 H+ N3- Initial (M) .05 Change (M) -x +x Final
Change
(M) -x +x
Final
.05-x X

6. Part B (continued)
As we figured out in part (a), the Ka can be written as [H+ ][N3-]/[HN3]
Based off our ICE table, this is equivalent to x2/(.05-x).
According to the problem, the value of Ka is 2.8e-5
This gives us the equation x2 /(.05-x)=2.8e-5. When we solve this, we find that x is According to our ICE table, x is equivalent to the concentration of [H+]. Therefore, the pH of the solution is –log(.0012), or 2.92.

7. Part C
We are given the amount of NaN3 in grams. Our first step will be to change this value into moles.
The molar mass of NaN3 is g. Therefore, .80 grams of NaN3 is equivalent to .012 moles of NaN3.
We know that there are .150 L of solution. Therefore, the concentration of NaN3, or the molarity, is equal to .012 moles/.150 L, or .082 M NaN3. Since there is one mole of the ion N3- for every mole of NaN3, the concentration of N3- will also be .082 M

8. Part C (continued)
Our next step is to write the equation describing the reaction that will occur when the NaN3 is added.
HN3 ↔ H+ + N3-
Side note: As we know, NaN3 and HN3 form a buffer solution, being as a buffer solution is a solution composed of a weak acid and the salt of its conjugate base.

9. Part C (continued)
To solve this problem, we again create an ICE table.
Our ICE table looks similar to the one constructed in (b), except in this case the initial concentration of N3- is The final concentration of N3- is now .082+x.

10. Part C (continued) HN3 H+ N3- Initial (M) .05 .082 Change (M) -x +x
.082
Change
(M) -x +x
Final
.05-x X
.082+x

11. Part C (continued) Ka can again be written as [H+ ][N3-]/[HN3]
Based off our ICE table, this is equivalent to (.082+x)x/(.05-x).
We set this equal to the value of Ka given in the problem, 2.8 x 10-5.
We find that x equals 1.7 x As x equals [H+ ], it follows that the pH of the solution is –log(1.7 x 10-5), or 4.77.

12. Part D We are told that .075 L of .1 M NaOH is added.
First, we need to convert this into moles by multiplying the molarity by the volume.
We find that 7.5 x 10-3 mols of NaOH are added, meaning essentially 7.5 x 10-3 moles of OH- are added as NaOH, being a strong acid, will dissociate completely into ions, and there are equal moles of NaOH and OH-.
We can similarly compute that there are 7.5 x 10-3 moles of HN3 initially. (.05 M x .150 L = 7.5 x 10-3 M)

13. Part D (continued)
Our next step is to write an equation describing the reaction that will take place.
OH-+HN3 ↔ H2O + N3-
We must attack this as a stoichiometric problem first, and then an equilibrium problem.
There are initially 7.5 x 10-3 moles of both HN3 and NaOH.
The moles of OH- will be completely consumed, and as it is in a one to one mole ratio with HN3, we can say that both will be completely consumed.
In other words, the final number of moles of both the reactants is 0. On the product side, both the H2O and the N3- increase by 7.5 x 10-3 moles.

14. Part D (continued) Now it is time to set up our ICE table.
Our ice table will be three columns by three rows. We label the first row Initial Concentration, the second row Change in Concentration, and the final row Equilibrium Concentration. We label the columns OH-, HN3, and N3- , the three aqueous substances involved in our chemical equation.
Because all the coefficients in the equation are 1, the mole to mole ratio of each substance is 1. Therefore, the changes that occur in equilibrium in the concentration of each substance are equal. The changes can be represented by the variable x. Because initially there is only N3- present, the N3- will decrease in concentration by x amount and both the OH- and the HN3 increase in concentration by x amount.

15. Part D (continued) We must find the initial concentration of the N3-.
We know from previous calculations that there are 7.5 x 10-3 moles of N3- after the reaction goes to completion.
To find molarity, we take the moles of solute divided by liters of solution.
In this case, the total volume of the solution is .15 L of the original solution added to .075 L of the NaOH solution, making a total of .225 L.
Therefore, the initial concentration of N3-, according to our calculations, is 3.33 x 10-2 M. The initial concentrations of the other substances are 0.

16. Part D (continued) OH- HN3 N3- Initial (M) 3.33 x 10-2 Change (M) +x
3.33 x 10-2
Change (M) +x -x
Equilibrium (M) X
3.33 x x

17. Part D (continued)
Based off our previous definition of Ka, we can find Kb in this reaction to be [OH-][HN3]/[N3-].
Based off our ICE table, this is equivalent to x2 /(3.33 x x).
We do not know the value of Kb; however, we know the value of Ka is 2.8 x Also, we know that the acid ionization constant of water, Kw, is equal to the products of Ka and Kb. Kw has a value of 1x
Therefore, by manipulating this information, we can realize that Kb equals 1x / 2.8 x 10-5, or 3.6 x
We produced the following equation: x2/(3.33 x x) = 3.6 x Solving for x, we find that x, or the concentration of OH-, is equal to 3.4 x 10-6.