Question

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Sol	Conc. NH3 (M)	Conc.NH4+(M)	pH
2	0.025 M	0 M	10.82
4	0.025 M	0.025 M	9.26
5	0.0025 M	0.0025 M	9.26

                                                                                                (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

N₁V₁ = N₂V₂   Where N₁ = conc. of added component.  N₂= final conc. of added component.

V₁ = volume of added component V₂ = final volume of solution

N₂ = N₁V₁ /V₂

Solution	[NH3] (M)	[NH4+] (M)	[NaOH] (M)	[HCl] (M)	pH
4	0.025	0.025	-	-	9.26
5	0.0025	0.0025	-	-	9.26
6	2.5*0.1/10=0.025	2.5*0.1/10=0.025	1*0.1/10=0.01	-	9.62
7	2.5*0.1/10=0.025	2.5*0.1/10=0.025	-	1*0.1/10=0.01	8.89
8	2.5*0.1/10=0.025	2.5*0.1/10=0.025	-	5*0.1/10=0.05	3.7
9	2.5*0.1/10=0.025	-	-	2.5*0.1/10=0.025	5.6

Comparing between 4 and 6, 6 has higher pH due to presence of additional NaOH as NaOH acts as base and increases pH.

If 0.01N NaOH was there without buffer then pH would be 14 + log (OH^-) = 14 + log (0.01) = 12

Solution 4 has the pH of 9.26 and upon addition of NaOH in solution 6, pH has not largely deviated (slight increased). So, it is clear that buffer reduced the change of pH, otherwise it would be as higher as 12 !

In presence of 0.05M HCl, 0.025M NH₃ got protonated and the residual solution has only 0.05M NH₄Cland excess 0.025M HCl. But, buffer acts when salt and acid both exists. Hence this solution 8 does not act as buffer.

Solution 9 has equal amount of acid and base. The residual solution has only NH₄Cl which is a salt of weak base and strong acid. This type of salt gives acidic solution upon hydrolysis. Hence pH is less than 7 i.e. in the acidic range.