Question

8. Calculate the pH for the following cases in the titration of 25.00 mL of 0.200-M acetic acid, CH3COOH(aq), with 0.200 M NaOH(aq): (a) before addition of any NaOH(aq) (b) after addition of 5.00 mL of NaOH(aq) (c) after addition of 12.50 mL of NaOH(aq) (d) after addition of 25.00 mL of NaOH(aq) (e) after addition of 26.00 mL of NaOH(aq)

Answer 1

Here we are dealing with a neutralization reaction taking place between acetic acid, CH3COOH, a weak acid, and sodium hydroxide, NaOH, a strong base.

Now, the pH of the resulting solution will depend on whether or not the neutralization is complete or not.

If the neutralization is not complete, We will have a buffer solution that will contain acetic acid and its conjugate base, the acetate anion..

The balanced chemical equation for this reaction is

CH3COOH(aq]+OH−(aq]→CH3COO−(aq]+H2O(l]

(a) beforw additon of any NaOH, only acidic acid be present having concentration 0.2 M

We can calculate the concentration of H+ ions in a 0.2 M solution of acetic acid by using the pKa or ionization constant which for any compound is the negative logarithm of the equilibrium coefficient of the neutral and charged forms.

pKa = -log(10)[Ka]

Ka for acetic acid is [H+][CH3COO-]/[CH3COOH] where [] indicates the concentration.

pKa = log(10)[CH3COOH] - log(10)[H+] - log(10)[CH3COO-]

as [H+] = [CH3COO-]

pKa = log(10)[CH3COOH] - 2*log(10)[H+]

For acetic acid pKa = 4.76 and -log(10)[H+] = pH

4.76 = log(10)(0.2) + 2*pH

=> 2*pH = 4.76 + 0.6990

=> pH = 5.459/2

=> pH = 2.7295

(b) after addition of 5.00 mL of NaOH (aq.)

1 mole of acetic acid will react with 1 mole of sodium hydroxide, shown here as hydroxide anions, OH−, to produce 1 mole of acetate anions, CH3COO−.

Use the molarities and volumes of the two solutions to determine how many moles of each you're adding

c=n/V⇒n=c*V

nacetic=0.20 M*25.00⋅10−3L=0.0050 moles CH3COOH

and

nhydroxide=0.20 M*5.00⋅10−3L=0.0010 moles OH−

Since we have fewer moles of hydroxide anions, the added base will be completely consumed by the reaction.

As a result, the number of moles of acetic acid that remain in solution will be

nacetic remaining=0.0050−0.0010=0.0040 moles

The reaction will also produce 0.0010 moles of acetate anions.

This means that we are now dealing with a buffer. Use the Henderson-Hasselbalch equation to find its pH

pH=pKa+log([conjugate base]/[weak acid])

Use the total volume of the solution to find the new concentrations of the acid and of its conjugate base

Vtotal=Vacetic+Vhydroxide

Vtotal=25.00 mL+5.00 mL=30.00 mL

The concentrations will thus be

[CH3COOH]=0.0040 moles/30.00*10−3L=0.1333 M

and

[CH3COO−]=0.0010 moles/30.00*10−3L=0.0333 M

The pKa of acetic acid is equal to 4.75.

The pH of the solution will thus be

pH=4.75+log(0.0333M/0.1333M)

= 4.15

Similarly rest part can be solved.