A 40.0 mL sample of a 0.30 M solution of NH3 is titrated with a 0.50 M HCl solut

Question

A 40.0 mL sample of a 0.30 M solution of NH3 is titrated with a 0.50 M HCl solution.

Calculate the pH of the solution when the following percentages of the HCl solution required to reach the equivalence point have been added.

Plot the data you determined for the pH at various points in the titration. What is the pH of the ammonia solution before the titration begins (0% of the required HCl added)?

2% of the total volume of HCl?

5% of the total volume of HCl?

20% of the total volume of HCl?

35% of the total volume of HCl?

50% of the total volume of HCl?

65% of the total volume of HCl?

90% of the total volume of HCl?

Explanation / Answer

There is no need to calculat ethe way you are doing calculation in the 2-90%

this is a buffer, since there will be:

NH3 and NH4+

B(aq) + H2O(l) <-> BH+(aq) + OH-(aq)

Weak base = B;

Conjugate acid = BH+

Neutralization of OH- ions:

BH+(aq) + OH-(aq) <-> B(aq) + H2O(l); in this case, OH- is neutralized by BH+, as well as B is created

Neutralization of H+ ions:

B(aq) + H+(aq) <-> BH+(aq)

Therefore:

calculate via:

mmol of base left = (100 - %) / 100

mmol of conjugat eformed = %/100

pH = pKa + log(NH3/NH4+)

pKa for NH4+ = 9.25

a) for 2 %

pH = pKa + log(NH3/NH4+)

pH = 9.25+ log((100-2) / 2))

pH = 10.94

b)

for 5 %

pH = pKa + log(NH3/NH4+)

pH = 9.25+ log((100-5) / 5))

pH = 10.53

c)

for 20%

pH = pKa + log(NH3/NH4+)

pH = 9.25+ log((100-20) / 20))

pH = 9.85

d)

for 35 %

pH = pKa + log(NH3/NH4+)

pH = 9.25+ log((100-35) / 35))

pH = 9.52

e)

for 50 %

pH = pKa + log(NH3/NH4+)

pH = 9.25+ log((100-50) / 50))

pH = 9.25

f)

for 65 %

pH = pKa + log(NH3/NH4+)

pH = 9.25+ log((100-65) / 65))

pH = 8.98

g)

for 90 %

pH = pKa + log(NH3/NH4+)

pH = 9.25+ log((100-90) / 90))

pH = 8.29

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.