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?
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.
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.
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
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
a. To write the net ionic equation for the reaction of the solution with a strong acid, we first need to identify the relevant species present in the solution.
The solution is made up of 1.00 M NH3 and 0.500 M (NH4)2SO4. NH3 is a weak base and will react with a strong acid to form NH4+ ions. (NH4)2SO4 is a salt that dissociates into NH4+ and SO4^2- ions in solution.
The strong acid being added will likely be HCl, which will dissociate completely into H+ and Cl- ions in solution.
Combining all the necessary ions, the net ionic equation for the reaction can be written as follows:
NH3 + H+ -> NH4+
b. To write the net ionic equation for the reaction of the solution with a strong base, we need to identify the relevant species present in the solution.
The solution is made up of 1.00 M NH3 and 0.500 M (NH4)2SO4. NH3 is a weak base, and adding a strong base will not cause any significant change in the solution since NH3 is already present.
c. To determine the moles of NH3 and NH4+ present in the reaction system before and after adding HCl, we need the initial volumes of the solutions and their concentrations.
Given:
Volume of the solution = 100 mL
Volume of HCl added = 10 mL
Concentration of HCl = 1.00 M
To calculate the moles of a solute, we use the formula:
moles = concentration (M) × volume (L)
Before adding HCl:
Moles of NH3 = concentration × volume = 1.00 M × 0.100 L = 0.100 moles
Moles of NH4+ = 0 (since (NH4)2SO4 does not contain NH4+)
After adding HCl:
Moles of NH3 = 0.100 moles (no reaction with HCl)
Moles of NH4+ = concentration × volume = 1.00 M × 0.010 L = 0.010 moles (from the reaction NH3 + H+ -> NH4+)
To find the pH of the resulting solution, we need to consider the acidity of NH4+ in water.
NH4+ is the conjugate acid of the weak base NH3, and it will undergo hydrolysis to produce H3O+ ions.
Since NH4+ is a weak acid, we can assume it only partially dissociates in water. The equilibrium expression is:
NH4+ + H2O ⇌ NH3 + H3O+
The equilibrium constant expression can be written as:
Ka = [NH3] × [H3O+] / [NH4+]
However, NH3 is in excess, so we can assume its concentration remains constant.
Thus, we can simplify the expression to:
Ka = [H3O+]/[NH4+]
By manipulating the expression, we can solve for [H3O+]:
[H3O+] = Ka × [NH4+]
The value of Ka for NH4+ can be found in a table as it is an established equilibrium for weak acids.
Lastly, to find the pH, we use the formula:
pH = -log[H3O+]
By substituting the appropriate values for [H3O+], we can calculate the pH of the resulting solution.