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)

pKb=4.74

14-4.74=9.26

Ka=5.5*10^-10

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

pKb=14-10.82=4.76

Kb=1.75*10^-5

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

Diluting doubled the K value.

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

No, roughly the same

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+

pKa=10.42

Ka=3.82*10^-11

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 rating: 100% (QA)

Solution 1 Use the concentration of NH3 and the measured pH above to determine the Kb of NH3 show the ICE table and calculations In order to calculate the Kb of NH3 we will first need to set up an ICE ... View the full answer