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Lab Activity: Strong Acids & Strong Bases, Weak Acids & Weak Bases,
and the Hydrolysis of Salts
Strong acids and strong bases completely ionize in water to make hydronium or hydroxide ions. Because the reactions essentially result in 100% ionization, the [H 3 O+ ] and the [OH−] can be calculated directly from the initial concentrations of the acid or base. Consider nitric acid, HNO 3 , and potassium hydroxide, KOH.
HNO 3 + H 2 O Æ H 3 O+^ + NO 3 −^ KOH Æ K+^ + OH−
[H 3 O+^ ]soln = [HNO 3 ] 0 [OH−]soln = [KOH] 0
On the other hand, weak acids and bases do not fully dissociate or ionize when they are in water. The pH of a solution of a weak acid or weak base must be predicted based on their dissociation constants: Ka or Kb. For an acid dissociation, consider hypochlorous acid, HClO.
HClO + H 2 O Ù H 3 O+^ + ClO 2 −
This reaction does not go to completion, therefore [H 3 O+^ ]soln ≠ [HClO] 0. Therefore, we must apply the equilibrium condition and use Ka to calculate the [H 3 O+^ ] of the solution. The pH can then be calculated as the −log[H 3 O+^ ]
A weak basic solution, like methylamine will not fully ionize water into hydroxide ions, but instead, an equilibrium is established:
CH 3 NH 2 + H 2 O Ù CH 3 NH 3 +^ + OH−
The [OH−]soln ≠ [CH 3 NH 2 ] 0. The hydroxide ion concentration must be calculated using Kb and the application of the equilibrium condition. The hydronium ion concentration and the pH can then be calculated.
Some salts exhibit acidic or basic properties based on the identities of their cations or anions. Other salts are completely neutral when dissolved in water. Consider KNO 3. When KNO 3 is dissolved in pure water, the pH of the solution will be neutral (7). This is because K+ is the conjugate acid of a very strong base, KOH. Therefore, the following reaction simply does not occur:
K+^ + H 2 O KOH + H+
Because NO 3 −^ is the conjugate base of a very strong acid, it also does not react with water:
NO 3 −^ + H 2 O NO 3 −^ + H 2 O
However, when the anion in a salt is the conjugate base of a weak acid, then the ion is itself a weak base, and will ionize slightly in water to make the solution basic. Consider potassium hypochlorite, KClO. As we have already discussed, potassium ions do not affect the pH of a solution. However, chlorite is the conjugate base of hypochlorous acid. Therefore, hypochlorite should be a weak base:
ClO−^ + H 2 O Ù HClO + H 2 O
Kb for the above reaction can be calculated from the Ka for chlorous acid:
Ka • Kb = Kw Kb = Kw / Ka = 1 x 10−^14 / 3.5 x 10−^8 = 2.86 x 10−^7
When the cation in a salt is the conjugate acid of a weak base, then the ion is a weak acid, and will dissociate slightly in water to make the solution acidic. Consider the salt methylammonium nitrate (CH 3 NH 3 NO 3 ). As already discussed, nitrate ions do not affect the pH of a solution. However, methylammonium ions are the conjugate acid of the weak base, methylamine. Methylammonium is a weak acid:
CH 3 NH 3 +^ + H 2 O Ù H 3 O+^ + CH 3 NH 2
Ka = Kw / Kb = 1 x 10−^14 / 4.6 x 10−^4 = 2.2 x 10−^11
When a salt is composed of a weak acid and a weak base, then you must compare the Ka and the Kb of the two ions to determine which will dominate. In the case of the salt methylammonium hypochlorite, CH 3 NH 3 ClO, Kb(ClO−) > Ka(CH3NH3+). Therefore, a solution of this salt should be basic.
In some cases, a cation or anion may be amphiprotic – it can accept or donate a proton to solution. An example of such an amphiprotic salt is potassium dihydrogen phosphate, KH 2 PO 4. Potassium undergoes no reaction with water. However, the dihydrogen phosphate has two possible reactions with water:
H 2 PO 4 −^ + H 2 O Ù HPO 42 −^ + H 3 O+^ Ka = 6.2 x 10−^8
H 2 PO 4 −^ + H 2 O Ù H 3 PO 4 + OH−^ Kb = 1.3 x 10−^12
Both reactions occur. However, because the Ka value is much larger than the Kb value, the acid dissociation dominates and the pH of the resulting solution is expected to be acidic. The pH can be predicted on the basis of the first reaction alone (because Ka>>Kb).
Some metal ions are slightly acidic or basic when dissolved in water because they form hydrated complexes that may then gain or lose protons from water. Consider the cobalt (II) ion in water. It produces a slightly acidic solution.
Co(H 2 O) 6 2+^ + H 2 O Ù Co(H 2 O) 5 OH+^ + H 3 O+^ Ka = 1.3 x 10−^9
PROCEDURE
For this activity, you will study a number of solutions to predict and analyze their pH. In some cases, you will prepare the solutions yourself. In others, you will simply take prepared solutions from a bottle. In each case:
- Predict the pH of the solution by calculation – considering the appropriate reaction with water and the Ka or Kb for the acid or base.
- Collect or prepare the solution.
- Measure the pH with a pH paper to get a preliminary pH reading.
- Calibrate a pH meter for the appropriate acid (4-7) or base (7-10) range.
- Measure the pH of the solution using the pH meter.
System #5: 0.10 M acetic acid (HC 2 H3 O 2 )
Reaction with water:
pH Calculation:
Predicted pH: __________ pH paper: _________ pH meter: _________
System #6: 0.10 M ammonia (NH 3 )
Reaction with water:
pH Calculation:
Predicted pH: __________ pH paper: _________ pH meter: _________
System #7: 0.10 M sodium acetate
Reaction with water:
pH Calculation:
Predicted pH: __________ pH paper: _________ pH meter: _________
System #8: 0.10 M ammonium chloride
Reaction with water:
pH Calculation:
Predicted pH: __________ pH paper: _________ pH meter: _________
System #9: 0.10 M ammonium acetate
Reaction with water:
pH Calculation:
Predicted pH: __________ pH paper: _________ pH meter: _________
System #10: 0.10 M sodium carbonate
Reaction with water:
pH Calculation:
Predicted pH: __________ pH paper: _________ pH meter: _________
Post-Lab Questions: Intro to Acids, Bases, & Salts
1. Compare the observed pH values for sodium bicarbonate and sodium carbonate. In terms of
Ka and Kb , explain why these differ.
2. Consider your calculation for System #5, what percent of the acetic acid is dissociated?
3. What would be the predicted pH of a 0.010 M acetic acid solution? What percent of the acid
is dissociated?
4. The solution in question #3 is a 10-fold decrease in the concentration of the acid vs. 0.10 M acetic
acid. Consider the following solutions of hydrochloric acid: 0.10 M HCl and 0.010 M HCl.
A. What are their predicted pH’s? 0.10 M HCl = __________ 0.010 M HCl = _________
B. Why do the HCl solutions have higher pH’s?
C. Why is there a more significant change in the pH of the two HCl solutions than in the two
acetic acid solutions?
