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?
6 answers
Jason, have you worked with the Henderson-Hasselbalch equation?
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.
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.
Ok, thank you! this makes a lot of sense. and the pKa is the -logKa? Ka=[NH4^+]=110?
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
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
1 answer