A solution of volume 0.500 L contains 1.70g NH3 and 4.15g (NH4)2SO4. What is the pH of this solution?
Answer obtained was 9.76, however it states that it is slightly off...
I get 9.76 also using 1.8E-5 for Kb. Use the Kb found in your text or notes. Watch the number of significant figures and rounding.
To calculate the pH of a solution, we need to know the concentration of the hydronium ion (H3O+). In this case, we have a mixture of NH3 and (NH4)2SO4. NH3 is a weak base and (NH4)2SO4 is a salt derived from a weak acid (NH4+) and a strong base (SO4^2-).
Let's calculate the concentration of the hydronium ion (H3O+) using the given information.
Step 1: Calculate the number of moles of NH3 and (NH4)2SO4.
To find the number of moles, we will use the formula: Moles = mass/molar mass.
Molar mass of NH3 = 14.01 g/mol + 1.01 g/mol x 3 = 17.03 g/mol.
Moles of NH3 = 1.70 g / 17.03 g/mol = 0.0999 mol.
Molar mass of (NH4)2SO4 = (14.01 g/mol + 1.01 g/mol) x 2 + 32.06 g/mol + 16.00 g/mol x 4 = 132.14 g/mol.
Moles of (NH4)2SO4 = 4.15 g / 132.14 g/mol = 0.0314 mol.
Step 2: Calculate the total volume of the solution.
The total volume of the solution is given as 0.500 L.
Step 3: Calculate the molarity of NH3 and (NH4)2SO4.
Molarity (M) = moles/volume.
Molarity of NH3 = 0.0999 mol / 0.500 L = 0.200 M.
Molarity of (NH4)2SO4 = 0.0314 mol / 0.500 L = 0.063 M.
Step 4: Write the dissociation equations for NH3 and (NH4)2SO4.
NH3 + H2O <-> NH4+ + OH-
(NH4)2SO4 <-> 2 NH4+ + SO4^2-
Since NH3 is a weak base, it will react with water to produce ammonium (NH4+) and hydroxide (OH-) ions. Since (NH4)2SO4 is a salt, it undergoes complete dissociation to produce ammonium (NH4+) and sulfate (SO4^2-) ions.
Step 5: Calculate the concentration of OH- ions.
Since the concentration of ammonium (NH4+) ions is twice that of NH3, the concentration of OH- ions can be calculated from the stoichiometry of the reaction:
[OH-] = [NH4+] = 2 x 0.063 M = 0.126 M.
Step 6: Calculate the concentration of H3O+ ions.
In a neutral solution, the concentration of H3O+ ions equals the concentration of OH- ions: [H3O+] = [OH-] = 0.126 M.
Step 7: Calculate the pOH.
pOH = -log([OH-])
pOH = -log(0.126)
pOH = 0.899
Step 8: Calculate the pH.
pH = 14 - pOH
pH = 14 - 0.899
pH = 13.101
The calculated pH of the solution is 13.101.