Unit 8: Topic 3 - Weak Acid and Base Equilibria

Question 1

What are weak acids?

Answer 1

• Most acids are weak; this means that unlike strong acids, weak acids will not fully dissociate into ions(hydronium ions and its conjugate base) inside of an aqueous solution.
• In other words, only a small percentage of the molecules of the weak acid will be ionized, with the majority remaining unionized.
• Strong acids include HCL, HNO3, H2SO4, HBr, HI, HClO4, and HClO3(memorize this list!). Any acid that is not on this list is considered a weak acid.

Question 2

What is the general equation for weak acids, and what are some of its implications?

Answer 2

• The general reaction for a weak acid at equilibrium in water is: HA(aq) + H2O(l) <–> H3O+(aq) + A-(aq) or more simply HA(aq) <–> H+(aq) + A-(aq), where HA represents a weak acid and A- represents the conjugate base of HA.
• The equilibrium constant for this reaction, K, is often referred to as the acid dissociation constant or acid ionization constant, denoted by Ka. It is also common to see the equilibrium constant being reported as pKa(pKa = -log Ka).
• Following the equilibrium reaction above, Ka = ([H3O+] * [A-]) / ([HA]).
• Since only a small percentage of the weak acid will dissociate, the equilibrium concentration of H+ ions is much less than the initial concentration of the weak acid. This means that the equilibrium lies far to the left and that the acid dissociation constant is often very small.

Question 3

How do you find the pH of a weak acid solution given the initial concentration of the weak acid and the acid dissociation constant?

Answer 3

Recall that pH = -log [H3O+] or simply pH = -log [H+]. This means that we need to find the equilibrium concentration for H+ in order to calculate the pH of the solution. To do this, we can use a RICE table, where x represents the equilibrium concentration of H+ and Ca represents the initial concentration of the weak acid. Note that [H+] = [A-] at equilibrium.
Reaction - HA(aq) <–> H+(aq) + A-(aq)
Initial - Ca 0 0
Change - -x +x +x
Equilibrium - Ca - x x x

Therefore, Ka = (x^2) / (Ca - x). However, we can assume that the equilibrium concentration of the acid will essentially be equivalent to the initial concentration because the weak acid would have gone through little ionization as the reaction goes towards equilibrium.
- Ka = x^2/Ca and after some manipulations, x = [H+] at equilibrium = sqrt(Ka * Ca).
- Finally we can now plug in the value of [H+] into the equation pH = -log [H+] in order to find the pH of the solution.
- Note: If you’re given pKa, simply convert it to Ka by computing Ka = 10^(-pKa)

Question 4

Calculate the pH of 2.0 M acetic acid HC2H3O2 with a Ka of 1.8*10^(-5).

Answer 4

Recognize that acetic acid will exist in equilibrium with its conjugate base and hydronium ions when reacted with water. Thus, its (simplified) acid dissociation equation would be:
HC2H3O2(aq) <–> H+(aq) + C2H3O2(aq)

Now we can calculate [H+] at equilibrium with a RICE table, with the initial concentration being 2.0 M:
Reaction - HC2H3O2(aq) <–> H+(aq) + C2H3O2(aq)
Initial - 2.0 0 0
Change - -x +x +x
Equilibrium - 2.0 - x x x

Therefore, Ka = (x^2) / (2.0 - x). However, we can assume that the amount of ionization that acetic acid will undergo will be negligible.
Thus, Ka = (x^2) / 2 = 1.8 * 10^(-5)
x = [H+] = 6.0 * 10^(-3)
pH = -log [H+] = -log( 6.0 * 10^(-3) ) = 2.22

Question 5

What are weak bases?

Answer 5

• Most bases are weak; this means that unlike strong bases, weak bases will not fully dissociate into ions(hydroxide ions and its conjugate acid) inside of an aqueous solution.
• In other words, only a small percentage of the molecules of the weak base will be ionized, with the majority remaining unionized.
• Elements from group 1(alkali metals) and group 2(alkaline earth metals) bonded with hydroxide are usually considered to be strong bases. Any base that is not on this list is considered a weak base.

Question 6

What is the general equation for weak bases, and what are some of its implications?

Answer 6

• The general reaction for a weak base at equilibrium is: B(aq) + H2O(l) <–> HB+(aq) + OH-(aq), where B represents a weak base and HB+ represents the conjugate acid of B.
• The equilibrium constant for this reaction, K, is often referred to as the base dissociation constant or base ionization constant, denoted by Kb. It is also common to see the equilibrium constant being reported as pKb(pKb = -log Kb).
• Following the equilibrium reaction above, Kb = ([HB+] * [OH-]) / [B]
• Since only a small percentage of the weak base will dissociate, the equilibrium concentration of OH- ions is much less than the initial concentration of the weak base. This means that the equilibrium lies far to the left and that the base dissociation constant is often very small.

Question 7

How do you find the pH of a weak base solution given the initial concentration of the weak base and the base dissociation constant?

Answer 7

Recall that pOH = -log [OH-]. This means that we need to find the equilibrium concentration of OH- in order to calculate the pOH of the solution, and therefore the pH of the solution. To do this, we can use a RICE table, where x represents the equilibrium concentration of OH- and Cb represents the initial concentration of the weak base. Note that [OH-] = [HB+] at equilibrium.
Reaction- B(aq) + H2O(l) <–> HB+(aq) + OH-(aq)
Initial- Cb 0 0
Change- -x +x +x
Equilibrium- Cb - x x x

Therefore, Kb = x^2/(Cb - x). However, we can assume that the equilibrium concentration of the weak base will essentially be equivalent to the initial concentration because the weak base would have gone through little ionization as the reaction goes towards equilibrium.
- Kb = x^2/Cb and after some manipulation, x = [OH-] at equilibrium = sqrt(Kb * Cb).
- We can now plug in the value of [OH-] into the equation pOH = -log [OH-] to find the pOH of the solution
- Recall that pKw = 14 = pOH + pH. Therefore, compute pH using pH = 14 - pOH.
- Note: If you’re given pKb, simply convert it to Kb by computing Kb = 10^(-pKb)

Question 8

Calculate the pH of a 0.150 M solution of NH3 with a Kb of .1.8 * 10^(-5).

Answer 8

Recognize that ammonia will exist in equilibrium with its conjugate acid and hydroxide ions when reacted with water. Thus, its base dissociation equation would be:
NH3(aq) + H2O(l) <–> NH4+(aq) + OH-(aq)

Now we can calculate [OH-] at equilibrium with a RICE table, with the initial concentration being 0.150 M:
Reaction - NH3(aq) + H2O(l) <–> NH4+(aq) + OH-(aq)
Initial - 0.150 0 0
Change - -x +x +x
Equilibrium - 0.150 - x x x

Therefore, Kb = (x^2) / (0.150 - x). However, we can assume that the amount of ionization that ammonia will undergo will be negligible.
Thus, Kb = x^2 / 0.150 = 1.8 * 10^(-5)
x = [OH-] = 1.64 * 10^(-3)
pOH = -log [OH-] = -log ( 1.64 * 10^(-3) ) = 2.78
pH = 14 - pOH = 14 - 2.78 = 11.22

Question 9

How do you calculate the percent ionization of a weak acid(or base) given the initial concentration of the weak acid(or base) and the acid(or base) dissociation constant?

Answer 9

The formula to find the percent ionization of a weak acid or base is:
% dissociation = [H+] / [HA] * 100% for weak acids and % dissociation = [OH-] / [B] * 100% for weak bases, where [H+] and [OH-] are equilibrium concentrations, and [HA] and [B] are initial concentrations.

To calculate [H+] and [OH-] at equilibrium, carry out an ordinary equilibrium calculation to obtain:
[H+] = sqrt(Ka * Ca) and [OH-] = sqrt(Kb * Cb), where Ca represents the initial concentration of the weak acid and Cb represents the initial concentration of the weak base. If you’re given the pKa or the pKb, simply convert them to Ka or Kb using the equations Ka = 10^(-pKa) and Kb = 10^(-pKb). These calculations assume that you’re dealing with a monoprotic weak acid or base.

Generally, the more dilute the weak acid or base is (lower initial concentration), the higher the percent dissociation will be.

Question 10

Calculate the percent ionization of a 0.20 M C2H5NH2 solution with a Kb of 5.6 * 10^(-4).

Answer 10

Recall that the percent dissociation for weak bases is % dissociation = [OH-] / [B] * 100%. In this case, [B] = [C2H5NH2] = 0.20 M. Next, recognize that ethylamine will exist in equilibrium with its conjugate acid and hydroxide ions when reacted with water. Thus, its base dissociation equation would be:
C2H5NH2(aq) + H2O(l) <–> C2H5NH3+(aq) + OH-(aq)

Now we can calculate [OH-] at equilibrium with a RICE table, with the initial concentration being 0.20 M:
Reaction - C2H5NH2(aq) + H2O(l) <–> C2H5NH3+(aq) + OH-(aq)
Initial - 0.20 0 0
Change - -x +x +x
Equilibrium - 0.20 - x x x

Therefore, Kb = (x^2) / (0.20 - x). However, we can assume that the amount of ionization that ethylmaine will undergo will be negligible.
Thus, Kb = x^2 / 0.20 = 5.6 * 10^(-4)
x = [OH-] = 1.06 * 10^(-2)
% dissociation = [OH] / [C2H5NH2] * 100% = ( 1.06 * 10^(-2) ) / ( 0.20) * 100% = 5.29%