Question

Sketch a pH titration curve if 100.0 mL of 0.125 M NH3 solution is titrated with 0.15M HCl. Note the following three items on the curve. K, for NHa 1.8 x 105 pH= Calculate the starting pH (no HCl added) vol (mL)= Calculate the volume of HCl added to reach the equivalence point pH = Calculate the pH at the equivalence point.

Answer 1

Given Molarity of NH3 = 0.125 M Volume of NH3 = 100.00 ml = 0.1000 L We know, Moles = Molarity (M) x Volume (V in L) Molarity of HCI= 0.15 M 1.Initial pH of solution NH3 is a weak Base , it dissociate partially into NH4+ and OH-in aqueous media as shown below NH3 (aq) + H2O <-----> NH4+ (aq) + OH- (aq) Initial 0.125 0 0 Change + x Equilibrium 0.125 - x - X Assume x is small then 0.125 - x will be = 0.125 M Given Kb= 1.80E-05 Weak acid dissociation constant Ka= (NH4+)(H+)/(NH3) 1.80E-05 = x*x=( 0.125 - x) 1.80E-05 = x*x: 0.125 x^2 0.00 0.125 SO X^2 = 2.25E-06 1.50E-03 but from ICE table x = [OH-] SO (OH) = 1.5E-03 M but pOH = -log [OH-] = -log[ 0.00150 ] = 2.8239 SO POH of buffer solution = 2.82 pH = 14- pOH = 14.000 - 2.824 11.18

4. pH at equivalent point Moles NH3 = 0.125 * 0.100 mol = 0.01250 mol Molarity of HCI= 0.150 M Given Ka= 1.8* 10^-5 we know pka = -log (Ka) = -log (1.8 * 10^-5) = 5-log(1.8)= 4.744 At equivalent point initial moles of NH3 will be completely nutralized and same amount of NH4+ will be formed Initial moles of NH3= 0.01250 mol; Moles of HCl needed to reach equivalent= 0.01250 mol volume of NH3= 100.0 ml; Molarity of HCl given = 0.125 M = 0.10000 L Volume of HCl needed to reach equilibrium = mol / Molarity = 0.01250 mol / 0.125 0.10000 L mol L-1 Total volume of solution at equivalent point = 0.10000 + 0.1000 0.2000 L When OH- is added to acetic acid, it reacts with acetic acid. and produce aceate. the ICE table is shown below. NH3 (aq) + H2) (aq) <-----> NH4+ (aq) + OH- (aq) Initial 0.0125 0.01250 0.000 Change -0.0125 -0.0125 + 0.01250 Equilibrium 0 0.01250 0.0 = 0.200000 L Above ICE table shows only presence of NH4+ = 0.01250 mol Concnetration of Acetate NH4+ = moles / total volume = 0.01250 mol = 0.0625 M Now the pH of acetate can be calculated as shown below Initial Change Equilibrium NH4+ (aq) 0.0625 ( x ) (0.0625-x) + H2O (aq) <-----> NH3 (aq) 0 ( +x ) (x) + H3O+ (aq) 0.000 (+x) (x) so Given Ka of NH3 = 1.8 x 10^-5 ; but Ka x Kb = Kw = 1.0 x 10^-14 Then Kb = Kw / Ka = 1.0 x 10^-14 / 1.8 x 10^-5 = 5.55 x 10^-10 Acetate is a weak base and equilibrium constant Kb can be expressed as below Kb = (NH3 x OH-) = (C3H302-) 5.55E-10 = (x) (x) = (0.0625-x) asume x is small then 5.55E-10 x^2 = 0.0625 (0.0625-x) will be 0.0625 5.55E-10 x 0.0625 sox^2 3.469E-11 5.890E-06 but from ICE table x = [H3O+] so [H3O+] = 5.89E-06 but pH = -log [H+] = -log [ 5.89E-6] = 5.2299 x^2 pH of solution at equivalent point = 5.23

volume of HCl =0.1 L = 100 ml