A 200.0 mL sample of 0.216 M strong acid is titrated with 0.170 M strong base.  Determine the pH of the solution before the addition of any strong base.

Chemistry & Chemical Reactivity
10th Edition
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter17: Principles Of Chemical Reactivity: Other Aspects Of Aqueous Equilibria
Section17.3: Acid-base Titrations
Problem 17.6CYU: The titration of 0.100 M acetic acid with 0.100 M NaOH is described in the text. What is the pH of...
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A 200.0 mL sample of 0.216 M strong acid is titrated with 0.170 M strong base.  Determine the pH of the solution before the addition of any strong base.

Expert Solution
Step 1

Strong acid – It is defined as an acid completely dissociate in aqueous solution called as strong acid.It has large capacity to give protons (H+) in aqueous solutions.

e.g. HCl, H2SO4, HNO3, HBr, HClO4 , HI

Weak acid- an acid dissociate slowly into its ions in an aqueous solution to give H+ ions

e.g. acetic acid.

Strong base- a base dissociate immediately into its ions in an aqueous solution to give OH- ions.

e.g. Sodium hydroxide.

Weak base-a base dissociate slowly into its ions in an aqueous solution to give OH- ions.

e.g. Ammonium hydroxide

At equivalence point of titration following three conditions can be observed-

  • Titration between strong acid and strong base results in neutral solution.
  • Titration between strong acid and weak base results in acidic solution.
  • Titration between weak acid and strong base results in basic solution.
Step 2

Answer of given numerical:

Consider,

Strong acid HA dissociate as follows and ICE table is 

      HA(aq)  + H2O(l) H3O+(aq) +A-(aq)I     0.216         -              0                  0C   -0.170      -            0.216       0.216  E       0              -            0.216      0.216

Where,I=Initial concentration,

C=Change in concentration,

E =Concentration at equilibrium

So,

[H3O+] =0.216 M

pH=-log [H3O+]pH=-log(0.216)pH=0.67

 

 

 

 

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