Question
A solution made up of 1.00 M NH3 and 0.500 M (NH4)2SO4 has a pH of 9.26.
a. Write the net ionic equation that represents the reaction of this solution with a strong acid.
b. Write the net ionic equation that represents the reaction of this solution with a strong base.
c. To 100. mL of this solution, 10.0 mL of 1.00 M HCl is added. How many moles of NH3 and NH4+ are present in the reaction system before and after the addition of the HCl? What is the pH of the resulting solution?
a. Write the net ionic equation that represents the reaction of this solution with a strong acid.
b. Write the net ionic equation that represents the reaction of this solution with a strong base.
c. To 100. mL of this solution, 10.0 mL of 1.00 M HCl is added. How many moles of NH3 and NH4+ are present in the reaction system before and after the addition of the HCl? What is the pH of the resulting solution?
Answers
DrBob222
Jason, have you worked with the Henderson-Hasselbalch equation?
jason
pH=pKa+ log(conjugate base/conjugate acid)
I have it in my notes, I just can't figure out the steps to solve this problem, or how it applies.
I have it in my notes, I just can't figure out the steps to solve this problem, or how it applies.
DrBob222
The base is NH3.
The acid is NH4^+.
a. With a strong acid it's the base that uses it; i.e.,
NH3 + H^+ ==> NH4^+
b. With a strong base it's the acid that uses it.
NH4^+ + OH^- ==> NH3 + H2O
c. So we start with 100 mL of the buffer.
millimols NH3 = mL x M = 100 x 1M = 100
mmols NH4^+ = mL x M = 100 x 0.5M x 2 = 100
mmols HCl added = 10mL x 1M = 10.
........NH3 + H^+ ==> NH4^+
I.......100...0........100
add...........10...........
C.......-10..-10........+10
E........90....0........110
pH = pKa + log (base)/(acid_
pH = pKa + log(90/110)
Plug in pKa and solve for pH.
The acid is NH4^+.
a. With a strong acid it's the base that uses it; i.e.,
NH3 + H^+ ==> NH4^+
b. With a strong base it's the acid that uses it.
NH4^+ + OH^- ==> NH3 + H2O
c. So we start with 100 mL of the buffer.
millimols NH3 = mL x M = 100 x 1M = 100
mmols NH4^+ = mL x M = 100 x 0.5M x 2 = 100
mmols HCl added = 10mL x 1M = 10.
........NH3 + H^+ ==> NH4^+
I.......100...0........100
add...........10...........
C.......-10..-10........+10
E........90....0........110
pH = pKa + log (base)/(acid_
pH = pKa + log(90/110)
Plug in pKa and solve for pH.
Jason
Ok, thank you! this makes a lot of sense. and the pKa is the -logKa? Ka=[NH4^+]=110?
DrBob222
Not quite.
If you use Kb for NH3 = 1.8E-5 then pKb = -log Kb = about 4.74 and since
pKa + pKb = pKw = 14, then
pKa = 14-4.74 = 9.26. Your tables may give a different value for Kb NH3 but most show 1.8E-5 or 1.75E-5
If you use Kb for NH3 = 1.8E-5 then pKb = -log Kb = about 4.74 and since
pKa + pKb = pKw = 14, then
pKa = 14-4.74 = 9.26. Your tables may give a different value for Kb NH3 but most show 1.8E-5 or 1.75E-5
NELLY
what PH would mark the end-point of a weak acid having a ka value of 0.000005 assume the salt formed to be at a molar concentration of 0.05 at the end