20.00 ml of 1.50 m ammonia(pKa=4.75) is titrated with 1.00 M hydrochloric acid
What is pKa?
What is initial ph?
What is ph after addition of 10.00ml of HCl?
What is ph at half equivalence point?
What is volume of titrant required to reach equivalence point?
What ph do you expect at equivalence point?
What is total volume after addition of 50.00 ml titrant?
What is ph after addition of 50.00 ml titrant?

See below.

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        

D) The broadest group of organization is known as the domain.