Chemical equilibria: Problems

1. Acetylcoline, a neurotransmitter, is quantitatively converted to choline and acetic acid by the enzyme acethylcholinesterase. 15 ml of the neurotransmitter (pH 7.65) are incubated with the enzyme and at the end of the reaction the pH of the solution result to be 6.87. Assuming that there was no buffer in the reaction mixture, calculate the moles of acetylcoline.

pH 7.65 corresponds to 10^-7.65 = 2.23 x 10^-8 moles/l H⁺

pH 6.87 corresponds to 10^-6.87 = 1.34 moles x 10^-7/l H⁺

Thus the moles/l of H⁺ formed are

1.34 moles x 10^-7 - 2.23 x 10^-8= 1.117 x 10^-7 mole/l

By inserting this value in the equilibrium constant we can calculate the concentration of acetic acid at equilibrium:

Ka = [H⁺]²/[ACH]

[ACH] = [H⁺]²/ Ka = (1.26 x 10^-14)/1.74 * 10^-5= 0.72 x 10^-9

The initial concentration of ACH formed is then

0.72 x 10^-9 + 1.117 x 10^-7 = 1.11 x 10^-7

The moles of acetic acid formed in 15 ml are:

1.11 x 10^-7 x 1000/15 = 1.7 x 10^-9

which corresponds to the moles of neurotransmitter originally present.

2. The pka of the amino group of glycine is 9.3. Calculate, for a 0.1 M solution at pH 9, the fraction of glycine having its amino group protonated.

The equilibrium to consider is

-NH₃⁺ -NH₂ + H⁺

for which

pH = pKa - log [-NH₃⁺]/[-NH₂]

log [-NH₃⁺]/[-NH₂] = pKa-pH = 9.3-9 = 0.3

[-NH₃⁺]/[-NH₂] = antilog(.3) = 2

Let z the amount of amino group protonated, thus:

z/(0.1-z) = 2

z = 0.2 - 2z

3z = .2

z=.2/3 = 0.066

(0.066 x 100)/.1 = 66%

3. The constant of equilibrium for the reaction N₂0₄ (g) = 2NO₂ (g) is 0.14 at 25^oC. Calculate the fraction of N₂0₄ dissociated when the total pressure is 1.5 atm.

Let

The total pressure is then P_t = P₁ + P₂ = 1.5 atm and the equilibrium constant can be written as follows:

K_p = P₁²/P₂ = P₁²/(P_t - P₁ ) = P₁²/(1.5 - P₁) = 0.14

from which

P₁² + 0.14 P₁ - 0.21 = 0

This is the well known form of the quadratic equation (see mathematic section). The solution of this equation gives P₁ = 0.40 or P₁ = -0.53, since the pressure cannot be negative P₁ = 0.40. Thus at equilibrium P₁ = 0.40 and P₂ = 1.5 - 0.40 = 1.1. According to Dalton's law:

P₁ = x₁P_t and P₂ = x₂P_t

we can calculate that

x₁ = P₁/Pt = 0.4/1.5 = 0.27 and of course x₂ = 1-0.27 = 0.73

In order to calculate the fraction of N₂O₄ dissociated, suppose that initially there was 100 moles of N₂O₄. If d is the number of moles dissociated, at equilibrium we must have:

by inserting these terms in the mole fraction expression we can calculate d:

About 15% of N₂O₄ dissociates, corresponding to the fraction 15/100 = 0.15.

On some books you can find the following formula to quickly solve the above problem:

K_p = 4d²P_t/(1-d²)

Since this formula can be difficult to remember and it is valid only for reactions of the same kind of the one considered above, we preferred to illustrate the solution of the problem by applying the general simple laws of gases.

For the solution of this problem we solved a quadratic equation which is quite simple to handle, however in many technical problems cubic or quartic equation, very tedious to solve, can be encountered. We shall illustrate here an approximate method for the solution of the quadratic equation above which can be easily applied to cubic or more complex equations.

K_p = P₁²/(1.5 - P₁) = 0.14

Let's  assume (arbitrarily) that P₁ is negligible compared to 1.5 thus

if this value is substituted in the equation of K_p, we have

K_p = 0.45²/(1.5 - 0.45) = 0.19

the value obtained for K_p is significantly different from the true value (0.14). However, if we use the value of P₁ in the equation of  K_p we can improve our result:

When this new value for P₁ is used to calculate K_p we have

K_p = 0.38²/(1.5 - 0.38) = 0.128

This new value of K_{p still differs from the true value but less than the value calculated before}

Again let's use the new value for P₁ to get another approximate value for P₁:

Now we can calculate a new value for K_p

K_p = 0.395²/(1.5 - 0.395) = 0.141

which is practically identical to the true value of K_p thus meaning the the value of P1 is a good approximation of the true value. In fact the value obtained by using the solution of the quadratic equation is 0.40.

By making further iteration we have

We can conlude that the method of successive approximations leads at each iteration to values converging to the true value. When two succesive values differ negligibly, the final value is a satisfactory approximation of the the true value.

The method described is arithmetically very simple and can be very useful in those problems were cubic or quartic equation mus be solved (or in the case you do not remember the formula for the solution of a quadratic equation).

5. Consider again problem 4 and calculate the fraction of N₂0₄ which dissociates when by increasing the volume of the system of reaction the pressure falls to 1 atm.

First let's try to predict the effect produced by a decrease in pressure (or an increase in volume). Since during the reaction there is a variation of the number of mole the equilibrium should be influenced by a change in pressure. Let' s write the equilibrium constant in terms of mole fraction:

K_p = (x₁P_t)²/ x₂P_t = x₁²P_t/ x₂ = (x₁²/ x₂)P_t

As can be seen a decrease in pressure must result in an increase of (x₁²/x₂), which can be obtained with an increase of the fraction of N₂O₄ which dissociates.

In order to calculate the new fraction which dissociates we can express K_p in terms of x₁ and solve the quadratic equation (try also the approximate method as made in problem 4):

K_p = (x₁² / 1- x₁)P_t

from which results x₁ = .31 corresponding to d = 0.18 (see problem 4 for details about calculations).

6. Considering again the reaction N₂0₄ (g) = 2NO₂ (g) calculate the partial pressure of N₂0₄ (P₂) when the partial pressure at the equilibrium of NO₂ (P₁) is 0.23 atm.

K_p = (P₁² / P₂) = 0.14

K_p = (0.052 / P₂) = 0.14

P₂ = 0.052 / 0.14 = 0.38 atm

7. Solid ammonium hydrosulfide dissociates in ammonia and hydrogen sulfide according to the following equation:

NH₄HS = NH₃(g) + H₂S(g), K_p = 0.11

the value of K_p include the concentration of solid at equilibrium. Calculate the partial pressure of the two gases when some solids NH₄HS is added to a flask containing 0.50 atm of ammonia.

K_p = P_NH3 P_H2S

there are two source of ammonia, the ammonia already present and the ammonia liberated from the dissociation:

Pt_NH3 = 0.5 + P_NH3

According to the stoichiometry of the reaction, P_NH3 = P_H2S

Pt_NH3 = 0.5 + P_H2S

K_p = (0.5 + P_H2S) P_H2S = 0.5P_H2S + (P_H2S)²

0.5P_H2S + (P_H2S)² - K_p = 0

Solving the quadratic equation we obtain P_H2S = 0.17 and P_NH3 = 0.5 + 0.17 = 0.67 atm.
 

8. Calculate the solubility of CaF₂ (K_sp = 1.7 x 10^-10) in a solution of 0.1 M Ca(NO₃)₂.

CaF₂ = Ca⁺⁺ + 2F^-

K_sp = [Ca⁺⁺][F^-]² = 1.7 x 10^-5

the total concentration of Ca⁺⁺ should be:

Ca⁺⁺_nitrate + Ca⁺⁺_fluoride = 0.1 + Ca⁺⁺_fluoride

from the value of K_sp  and considering that the dissociation of CaF₂ is depressed according to the LeChatelier principle, it can be deduced that the concentration of Ca⁺⁺ coming from CaF₂ is small compared to 0.1 M  thus we assume that the total concentration of calcium is 0.1 M

K_sp = [0.1][F^-]² = 1.7 x 10^-10

[F^-]² = 1.7 x 10^-10/ 0.1

[F^-]= 1.7 x 10^-10/ 0.1 = 4.1 x 10^-5

since each mole of CaF₂ which dissociates give raise two moles of F^-, the solubility of calcium fluoride is

4.1 x 10^-5/2 = 2 x 10^-5 M

This result justifies our neglect of the Ca⁺⁺ coming from CaF₂.

9. A solution contain 0.1 M Cl^- and 0.001 M CrO₄^--. The K_ps for AgCl and Ag₂CrO₄ are:

K_psAgCl = 2.8 x 10^-10

K_psAg2CrO4 = 1.9 x 10^-12

If silver nitrate is slowely added to the solution, calculate the amount of Cl^- left in solution when Ag₂CrO₄ starts to precipitate.

From the value of K_ps we can calculate the concentration of Ag⁺ at which Ag₂CrO₄ starts to precipitate.
 

K_psAg2CrO4 = [Ag⁺]² [CrO₄^--] = [Ag⁺]² 0.01 = 1.9 x 10^-12[Ag⁺]²  = 1.9 x 10^-12/0.01

[Ag⁺]  = 1.4 x 10^-5

by inserting this value in the kps of AgCl we can calculate the concentration of chloride ion at equilibrium with the calculated concentration of Ag⁺:
 

K_psAgCl = [Ag⁺] [Cl^-] = 1.4 x 10^-5[Cl^-] = 2.8 x 10^-10

[Cl^-] = 2.8 x 10^-10/1.4 x 10^-5 = 2 x 10^-5

The concentration of Cl^- is very small compared to the original concentration (0.1M), thus the precipitation can be considered quantitative.

10. Calculate 1) the pH of a solution which is 0.15 M in HCl and 0.05 M in NaOH and 2) the pH of a solution which is 0.05 M in HCl and 0.15 M in NaOH.

This solution can be considered equivalent to a solution 0.1 M (0.15 - 0.05) in HCl thus

pH = -Log (0.1) = 1

In the same way a solution 0.05 M in HCl and 0.15 M in NaOH, can be considered equivalent to a solution 0.1 M in NaOH thus

pOH = -Log (0.1) = 1

pH = 14 - 1 = 13

11. Calculate the pH of a solution 1M in HCl and 1M in NaOH.

In this case the only source of H₃O⁺ is water dissociation thus pH = 7 as in pure water.

Note: The solution above is equivalent to a solution 1M in NaCl, i.e., the salt formed by the strong acid HCl and the strong base NaOH.

12. Calculate the pH of a solution obtained by mixing 500 ml of 0.1 M HCl and 500 ml of 0.05 M ammonia (NH₃).

500 ml of 0.1 M HCl contains 0.05 moles of the acid, after mixing the volume is 1 liter thus HCl concentration becomes 0.05 M. Analogously NH₃ concentration becomes 0.025 M. The reaction of neutralization can be written as:

NH₃+ H⁺ --> NH₄⁺

This reaction will produce 0.05 - 0.025 = 0.025 moles/liter of NH₄⁺ leaving 0.025 moles/liter of H⁺. Therefore:

pH = - log 0.025 = 1.6

13. Calculate the pH of one liter of solution prepared by mixing 1.3 moles of acetic acid (pKa 4.75) and 0.6 moles of NaOH.

[CH₃COO^-] = 0.6 M (the concentration of the base added)

[CH₃COOH] = 1.3 - 0.6 = 0.7 M

pH = pKa - Log [CH₃COOH]/[CH₃COO^-] =
4.75 - Log (0.7) + Log (0.6) = 4.68

14. Calculate the pH of one liter of solution containing 0.6 moles of sodium acetate and 0.7 moles of acetic acid.

This problem is perfectly equivalent to the one above. In fact, the dissociation of sodium acetate (strong electrolytes) produce 0.6 moles of acetate. Thus, we will have a solution containing 0.6 moles of acetate and 0.7 moles of acetic acid. In this case the equation used above assumes the following form:

pH = pKa - Log Ca/Cs

where Ca is the concentration of the acid and Cs the concentration of the salt. Thus

pH = 4.75 - Log (0.7) + Log (0.6) = 4.68