Answer:
See below.
Step-by-step explanation:
pH of 1.50M NH₃(aq)
NH₃ + H₂O => NH₄OH ⇄ NH₄⁺ + OH⁻; Kb = 1.8 x 10⁻⁵
pKa = -log(Ka) = -log(1.8 x 10⁻⁵) = 4.74
C(i) 1.5M 0.0M 0.0M
ΔC -x +x +x
C(eq) 1.5 - x ≅ 1.5M x x
Kb = [NH⁺][OH⁻]/[NH₄OH] = x²/1.5 = 1.8 x 10⁻⁵
x = [OH⁻] = √1.5(1.8 x 10⁻⁵) = 0.0052M
pOH = -log[OH⁻] = -log(0.0052) = 2.28
pH = 14 - pOH = 14 - 2.28 = 11.71
pH after adding 10ml 1M HCl to 20ml 1.5M NH₃
10ml(1M HCl) + 20ml(1.5M NH₃) => 10ml(1M HCl) + 20ml(1.5M NH₄OH)
=> 0.010(1.0) mole HCl + 0.20(1.5) mole NH₄OH
=> 0.010 mole HCl + 0.030 mole NH₄OH
=> (0.030 - 0.010)mole NH₄OH + 0.010 mole NH₄Cl
=> 0.020 mole NH₄OH + 0.010 mole NH₄Cl
=> (0.020mol/0.030L) NH₄OH + (0.010mol/0.030L) NH₄Cl
=> 0.67M NH₄OH + 0.33M NH₄Cl
NH₄OH ⇄ NH₄⁺ + OH⁻
C(i) 0.67M 0.33M 0.00M
ΔC -x +x +x
C(eq) ≅ 0.67M ≅ 0.33M x
Kb = [NH₄⁺][OH⁻]/[NH₄OH] = (0.33)(x)/(0.67) = 1.8 x 10⁻⁵
x = [OH⁻] = (1.85 x 10⁻⁵)(0.67)/(0.33) = 3.66 x 10⁻⁵M
pOH = -log[OH⁻] = -log(3.66 x 10⁻⁵) = 4.44
pH = 14 - pOH = 14 - 4.44 = 9.56
Volume Titrant to Equivalence Point
NH₄OH + HCl => NH₄Cl + H₂O
moles NH₄OH = moles HCl
moles = Molarity x Volume
M(NH₄OH) x Vol(NH₄OH) = M(HCl) x Vol(HCl)
(1.5M)(20ml) = (1.0M)(Vol(HCl))
Vol 1M HCl needed to reach eqv. pt. = (1.5M)(20ml)/(1.0M) = 30ml 1M HCl.
pH at V(eq)/2:
Volume 1M HCl at V(eq)/2 = 15ml
=> 15ml(1M HCl) + 20ml(1.5M NH₄OH)
=> 0.015(1.0)mole HCl + 0.020(1.5)mole NH₄OH
=> 0.015 mole HCl + 0.030 mole NH₄OH
=> (0.030 - 0.015) mole NH₄OH + 0.15 mole NH₄⁺ + 0.15mol Cl⁻
=> (0.015mol/0.045L) NH₄OH + (0.15mol/0.045L) NH₄⁺ (drop Cl⁻ as spec ion)
=> 0.33M NH₄OH + 0.33M NH₄+
Kb = [NH₄⁺][OH⁻]/[NH₄OH] = (0.33M)[OH⁻]/(0.33M) = 1.8 x 10⁻⁵
=> [OH⁻] = 1.8 x 10⁻⁵M
=> pOH = -log[OH⁻] = -log(1.8 x 10⁻⁵) = 4.75
=> pH = 14 - pOH = 14 - 4.75 = 9.25
Rxns at Equivalence Point:
NH₄OH + HCl => NH₄Cl + H₂O
NH₄Cl => NH₄⁺ + Cl⁻
Hydrolysis Rxns:
Cl⁻ + H₂O => No Reaction (Cl⁻ will not hydrolyze to HCl)
NH₄⁺ + H₂O ⇄ NH₄OH + OH⁻ (<= This rxn will occur)
pH at Eqv Pt:
20ml(1.5M NH₄OH) + 30ml(1.0M HCl)
=> 0.020(1.5)mol NH₄OH + 0.030(1.0)mol HCl
=> 0.030mol NH₄OH + 0.030mol HCl
=> 0.030mol NH₄Cl + 0.030mol H₂O (drop H₂O)
=> 0.030mol NH₄Cl/0.050L soln = 0060M NH₄Cl
=> 0.060M NH₄⁺ + 0.060M Cl⁻ (drop Cl⁻ as spec ion)
Hydrolysis of NH₄⁺:
NH₄⁺ + H₂O ⇄ NH₄OH + H⁺
C(i) 0.060M ------ 0.00M 0.00M
ΔC -x ----- +x +x
C(eq) ≅0.060M ----- x x
Ka(NH₄⁺) = Kw/Kb = [NH₄OH][H⁺]/[NH₄⁺] = x²/0.060M = 1 x 10⁻¹⁴/1.5 x 10⁻⁵
=> x = [H⁺] = √(0.060)(1 x 10⁻¹⁴)/(0.060) M = 1.83 x 10⁻⁵M
=> pH = -log[H⁺] = -log(1.83 x 10⁻⁵) = 4.74
pH after adding 50ml 1.0M HCl:
Total Volume = 20ml + 50ml = 70ml of solution
=> 20ml(1.5M NH₄OH) + 50ml(1.0M HCl)
=> Solution containing NH₄⁺ ions, Cl⁻ ions and H⁺ ions
The 1.0M HCl will be in excess by 20ml and will dominate the pH value. The amount of H⁺ from NH₄⁺ hydrolysis is negligible with respect to the excess HCl. Therefore, it is assumed that all H⁺ is from the 20ml excess if 1M HCl.
=> moles HCl = 0.020L(1.0M) HCl = 0.020 mole HCl in 70ml solution
=> [HCl] excess = (0.020mol/0.070L) HCl = 0.286M HCl excess
∴ pH after adding 50ml 1.0M HCl = -log(0.286) = 0.54