Question

Equilibria with Weak Acids & Weak Bases - Results

Exp	Composition	Total Vol	pH
1	A few mL of 0.10M NH3 (unchanged)	n/a	11.13
2	2.5mL 0.1M NH3 & 7.5mL H2O	10 ml	10.82
3	2.5mL 0.1M NH4Cl & 7.5mL H2O	10 ml	6.01
4	2.5mL 0.1M NH3 & 2.5mL 0.1M NH4Cl & 5mL H2O	10 ml	9.26
5	Take 1mL Solution #4 and add 9mL of water	10 ml	9.26

Solution 1. Use the concentration of NH₃ and the measured pH above to determine the Kb of NH₃. (show the ICE table and calculations)

          Use the equation (Ka*Kb = 1*10^-14) to calculate the Ka for NH₄⁺ ?

Solution 2 Use the equation M₁V₁ = M₂V₂ to calculate new conc. of NH₃.

Conc. of NH₃ ____         pH measured:    10.82

Compare solutions 1 & 2. What effect did diluting the base have on the pH?

• Decreased the pH of the solution.

Use the new concentration and pH to measure the Kb of NH₃ again. Did it change?

Solution 3    Conc. of NH₄⁺                                pH of solution 3: 6.01

Use the diluted concentration of NH₄⁺ and measured pH to determine the Ka of NH₄⁺

How does this Ka value compare to the value on pg. 1? (within a power of 10 is good)

Calculate the concentrations for solution 4 and solution 5

Sol	Conc. NH3 (M)	Conc.NH4+(M)	pH
2			10.82
4			9.26
5			9.26

Compare pH of solutions 2&4. What effect did adding NH₄⁺ have on the equilibrium/pH?

Compare the pH from dilution in solutions 1&2 to the pH change in solutions 4 & 5. Why is the pH change smaller for solutions 4&5 than in solutions 1&2? Think above the ratio of base and conjugate acid. You can also show this mathematically by using the Kb from part 1 with NH₃/NH₄⁺ conjugates to determine what the OH- concentration should be.

Note: Solution 4 (page 2) should already be made.                                                                                                                                           (total volume)

6	2.5mL 0.10M NH3, 2.5mL 0.1M NH4Cl & 1.0mL 0.10M NaOH & 4mL H2O	10 mL
7	2.5mL 0.10M NH3, 2.5mL 0.1M NH4Cl & 1.0mL 0.10M HCl of & 4mL H2O	10 mL
8	2.5mL 0.10M NH3, 2.5mL 0.1M NH4Cl & 5.0mL 0.10M HCl	10 ml
9	2.5mL 0.10M NH3, 2.5mL 0.10M HCl & 5.0mL H2O	10 ml

Calculate the dilution concentrations of solution 6.

Sol	Conc. NH3 (M)	Conc. NH4+ (M)	Conc. NaOH(M)	pH
6				9.62

Compare solutions 4 & 6. What effect did NaOH have on the pH of the solution?

If the NaOH concentration from solution 6 was in water by itself (without the buffer), what would the pH be?

Did NH₃ & NH₄⁺ buffer against NaOH? How can you tell?

Calculate the concentrations of solution 7 & 8

Sol	Conc. NH3 (M)	Conc. NH4+ (M)	Conc.HCl (M)	pH
7				8.89
8				3.7

Compare solutions 4, 7 & 8. The ingredients are essentially the same, but the solutions of 7 & 8 should have a dramatically different pH.   Why did 7 work as a buffer, but 8 did not? Compare the concentration of NH₃ vs the concentration of HCl.  

Calculate the concentrations of solution 9

Sol	Conc. NH3 (M)	Conc.HCl (M)	pH
9			5.6

a) This solution has equal amounts of acid and base. Why is the pH acidic? Explain.

Answer 1