Question

When creating a titration curve for a weak base, the pH of the initial solution requires setting up a table showing the initial, change, and equilibrium values for each species and plugging these into the equilibrium constant expression. The expression for the bicarbonate ion is Ka1=[H2CO3][H+][HCO3−] For an anion that can both hydrolyze and produce H+, the pH of a concentrated solution can be more easily approximated using the equation pH=12(pKa1+pKa2) During the titration before the equivalence point, provided that the concentration of acid is significantly more than the concentration of base, the Henderson-Hasselbalch equation can be used to approximate the pH: pH=pKa+log[base][acid] At the equivalence point, the solution is no longer a buffer, but contains the weak acid H2CO3. The change in concentration of the bicarbonate ion is significant, and the equilibrium constant expression for Ka1 must again be used to find the concentration of hydronium ions. After the equivalence point, the strong acid will control the pH.

Part A

A 10.0-mL sample of 1.0 M NaHCO3 is titrated with 1.0 M HCl (hydrochloric acid). Approximate the titration curve by plotting the following points: pH after 0 mL HCl added, pH after 1.0 mL HCl added, pH after 9.5 mL HCl added, pH after 10.0 mL HCl added (equivalence point), pH after 10.5 mL HCl added, and pH after 12.0 mL HCl added.

Answer 1

HCL(aq)+NH3(aq)---NH4+(aq)+H2O

(l) Calculate the pH at the following points in the titration when 25.0 mL of 0.0100 M NH3 is titrated with 0.100 M HCL.

A. 0.0 mL of HCl added

moles NH3 = 0.0250 L x 0.0100 M=0.00025 mol

Kb = 1.8 x 10^-5
pKb = 4.74

a)
NH3 + H2O <=> NH4+ + OH-
1.8 x 10^-5 = x^2/ 0.0100-x

Due to small value of Kb 0.0100-x= 0.0100

X^2= 1.8*10^-7

X=4.24*10^-4
x = [OH-]= 4.24*10^-4M
pOH = 3.37
pH = 14 – 3.37

= 10.63

B. 10.0 mL of HCl added of 0.100 M HCl

25.0 mL of 0.100 M NH3

moles NH3 = 0.0250 L x 0.100 M=0.0025 mol

moles HCl = 0.0100 L x 0.100 M= 0.001 mol HCl

this react with NH3 IN 1;1 ratio.
moles NH3 in excess = 0.00250 - 0.001=0.0015
moles NH4+ = 0.001
total volume =10.0 ml+25.0ml= 35 ml= 0.0350 L
[NH3]= 0.0015 / 0.0350=0.043 M
[NH4+]= 0.001/ 0.0350=0.0285 M
pKb = 4.76
pOH = 4.76 + log 0.0285/0.043

pOH = 4.76 + log 0.663

pOH = 4.76 -0.178

=4.582
pH=14-pOH=14-4.582

=9.418

C. At equivalence point

moles HCl required to reach the equivalence point = 0.0025

volume HCl = 0.0025/ 0.100 = 0.0250 L

25+25 = 50 ml or 0.0500 L
total volume = 0.0500 L
moles NH4+ formed = 0.00500
[NH4+]= 0.0250/ 0.0500=0.500 M

NH4+ <=> NH3 + H+
Ka = Kw/Kb = 5.6 x 10^-10 = x^2/ 0.500-x

5.6 x 10^-10 = x^2/ 0.500-x

x = [H+]= 1.67 x 10^-5 M

pH = 4.78

D. 35.0 mL of HCL added

moles HCl = 0.0350 x 0.100=0.0035 HCl mol
moles H+ in excess = 0.0035- 0.0025 = 0.001
total volume = 0.060 L
[H+]= 0.001/0.0600=0.0167 M

pH =- log [H+]= -log 0.0167

pH=1.78