Question

Please show all your work for the problems!

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?

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)

Answer 1

Solution 1. pOH = 1/2pKb - 1/2 log(c)

So, 14-11.13 = 1/2pKb - 1/2 log(0.1). Hence, pKb = 4.74

Thus, Kb of NH3 = 1.8 x 10^-5

pKa of NH4+ = 14 - 4.74 = 9.26

Thus, Ka of NH4+ = 5.4954 x 10^-10

Solution 2. Concentration of NH3 = (2.5*0.1)/10 = 0.025 M

On Dilution of the Base, the [OH-] dropped, and thus the pH also dropped.

Again, 14-10.82 = 1/2pKb - 1/2 log(0.025)

So, pKb = 4.76. Thus, the pKb doesn't change significantly.

Kb = 1.746 x 10^-5

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Solution 3. Concentration of NH4+ = (2.5*0.1)/10 = 0.025 M

Now, 6.01 = 1/2pKa - 1/2 log(0.025)

Thus, pKa = 10.42

Hence, Ka = 3.82 x 10^-11

Thus, the value compares good according to the Solution 1.