Ammonia is a weak base with a Kb of 1.8 x 10-5. A 100.0 mL sample of 0.13 M aqueous solution of ammonia (NH3) is mixed with 87 mL of 0.15 M solution of the strong acid HCl, calculate the pH of the final solution.
A) 2.9
B) 11.1
C) 8.8
D) 5.2
E) None of the above
millimols NH3 = mL x M = 100 x 0.1 = 13
mmols HCl = 87 x 0.15 = 13.05
My best guess is that these are to be considered the same since the 87 has two s.f. and 13 does too.
........NH3 + HCl ---> NH4Cl
I.......13....13........0
C......-13...-13.......+13
E........0.....0.........0
I'm assuming the HCl exactly neutralizes the NH3 in which case the pH is determined by the hydrolysis of the salt, NH4Cl, which has a concn of 13mmols/187 mL = about 0.07 and you need to clear up that number a little as well as all that folllow.
........NH4^+ + H2O ==> NH3 + H3O^+
I.......0.07............0......0
C.......-x..............x......x
E......0.07-x...........x......x
Ka for NH4^+ = (Kw/Kb for NH3) = (x)(x)/(0.07-x) and solve for x = (H3O^+ and convert to pH. NH4^+ salt have a pH of about 5.2
To calculate the pH of the final solution, we need to first determine the amount of NH3 and HCl that react and then calculate the concentration of OH- ions in the solution.
Step 1: Calculate moles of NH3 and HCl
Moles of NH3 = Volume (in L) x Concentration (in mol/L)
Moles of NH3 = (100.0 mL / 1000 mL/L) x 0.13 mol/L = 0.013 mol
Moles of HCl = Volume (in L) x Concentration (in mol/L)
Moles of HCl = (87 mL / 1000 mL/L) x 0.15 mol/L = 0.013 mol
Since the stoichiometric ratio between NH3 and HCl is 1:1, both have reacted completely.
Step 2: Calculate the concentration of NH3 and NH4+ ions in solution
Since NH3 is a weak base, it reacts with water to form NH4+ and OH- ions.
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)
Initial concentration of NH3 = 0.13 mol/L
Moles of NH3 reacting = 0.013 mol
Since NH3 is a weak base, we assume it completely reacts with water, resulting in an equal concentration of NH4+ and OH- ions formed.
Concentration of NH4+ and OH- ions = Moles of NH3 reacting / Total volume (in L)
Concentration of NH4+ and OH- ions = 0.013 mol / (100.0 mL + 87 mL) = 0.013 mol / 0.187 L = 0.0695 mol/L
Step 3: Calculate the pOH of the solution
pOH = -log10[OH-]
pOH = -log10(0.0695) = 1.16
Step 4: Calculate the pH of the solution
pH + pOH = 14
pH = 14 - pOH = 14 - 1.16 = 12.84
Therefore, the pH of the final solution is approximately 12.84.
The correct answer is not listed among the given options.