Question

Pre-Laboratory Questions (An Acid Base Titration)

1. Show the reaction between HCl and NaOH. Label the acid, base, and each of the products.

2. What occurs at the equivalence point?

3. Consider the titration of 10.00 mL of 0.10 M HCl with 0.10 M NaOH

a.) What salt is formed during this reaction?

b.) Do you expect the salt solution at the equivalence point to be acidic, neutral, or basic?

4. Consider the titration of 10.00 mL of 0.10 M acetic acid (CH3COOH) with 0.10 M NaOH

a.) What salt is formed during this reaction?

b.) Do you expect the salt solution at the equivalence point to be acidic, neutral, or basic?

5. Calculate the pH of the solution in question 4 at the equivalence point.

Answer 1

acid is HCl sicne it can loose H+. While base is NaOH.

NaOH+ HCl -----------> NaCl + H2O. NaCl is salt.

Both NaOH and HCl are strong. hene salt formed at the equivalence point ( at equivalence point moles of NaOH= moles of HCl)

the salt is neutral

4. The salt formed is CH3COONa. Sodium acetate CH3COOH+ NaOH----> CH3COONa + H2O

since acetic acid is weak acid and sodium hydroxide is basic, so it is a base.

at equivalence point, volume of sodium hydroxide = 10*0.1= 0.1*V

V= volume of sodium hydroxide =10ml

moles of sodium acetate formed = 0.1*10/1000 =0.001

volume of solution after mixing = 10+10=20ml =20/1000L= 0.02L

concentration of sodium acetate= 0.001/0.02 =0.05M

CH3COONa+ H2O-------> CH3COOH+OH-

Kb= 10^-14/ Ka

for acetic acid Ka= 1.78*10-5

Kb= 10-14/ 1.84*10-5 =5.6*10^-10

let x= drop in concentration of CH3COONa to reach equilibrium . [OH-] = [CH3COOH]

x²/(0.5-x) = 5.6*10^-10, when solved using excel , x= 0.0000167

pH= 4.78