Question

Find the pH of the equivalence point and the volume (mL) of 0.0372 M NaOH needed to reach the equivalence point in the titration of 42.2 mL of 0.0520 M CH3COOH.

Answer 1

Answer:

8.54

Explanation:

At equivalence point :  

42.2 X 0.052 = Vol. NaOH X 0.0372

Vol of NaOH = 2.1944/0.0372 = 58.99 ml

So volume of NaOH recquired to reach equivalence point = 58.99 ml

Number of miliimoles of CH3COOH = molarity X volume in ml = 42.2 X 0.052             = 2.1944 millimoles

Number of millimoles of NaOH = 58.99 X 0.0372 = 2.1944

Now CH₃COOH and NaOH reacts to give CH₃COONa according to the reaction :

CH₃COOH + NaOH ------> CH₃COONa + H₂O

1 mole of CH₃COOH reacts with 1 mole of NaOH to give 1 mole of CH₃COONa  

So 2.1944 millimoles of CH₃COOH will react with 2.1944 millimoles of NaOH to give 2.1944 millimoles of CH₃COONa

So all the acid (CH₃COOH) and base (NaOH) has been converted into salt (CH₃COONa) so there is no acid or base left.

Now molarity of CH₃COONa = number of millimoles of CH₃COONa/total volume in ml = 2.1944/(58.99 + 42.2) = 2.1944/101.19 = 0.02169 M

So using the hydrolysis equation :  

pH = 1/2 [ pKw + pKa + log c ]  

Ka for acetic acid = 1.75 X 10⁻⁵  

so pKa = -log (1.75 X 10⁻⁵) = 4.74  

Kw = 10⁻¹⁴

so pKw = -log 10⁻¹⁴ = 14

c = 0.02169  

so log c = log 0.02169 = -1.66  

putting the values....  

pH = 1/2 [14 + 4.74 - 1.66 ]  

pH = 1/2 [ 17.08] = 8.54

Answer 2

Answer:

58.99mL of 0.0372 M NaOH to reach equivalence point

pH = 8.54

Explanation:

Moles of 42.2 mL of 0.0520 M CH₃COOH are:

0.0422L × (0.0520mol / L) = 2.194x10⁻³ mol CH₃COOH

That react with NaOH, thus:

CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O

That means 1 mole of acetic acid reacts per mole of NaOH

Thus, moles of NaOH you need to reach equivalence point are 2.194x10⁻³ mol, in volume:

2.194x10⁻³ mol × (1L / 0.0372mol) = 0.05899L ≡ 58.99mL of 0.0372 M NaOH to reach equivalence point

In the equivalence point, you will have just CH₃COO⁻ that is in equilibrium with water, thus:

CH₃COO⁻(aq) + H₂O(l) ⇄ CH₃COOH(aq) + OH⁻(aq)

And equilibrium is defined by:

Kb = 5.6x10⁻¹⁰ = [CH₃COOH] [OH⁻] / [CH₃COO⁻]

Molarity of CH₃COO⁻ is: 2.194x10⁻³mol / (0.05899L + 0.0422L) = 0.02168M

Thus, concentrations in equilibrium are:

[CH₃COO⁻] = 0.02168M - X

[CH₃COOH] = X

[OH⁻] = X

Replacing in Kb:

5.6x10⁻¹⁰ = [X] [X] / [0.02168M - X]

1.214x10⁻¹¹ - 5.6x10⁻¹⁰X = X²

X² + 5.6x10⁻¹⁰X - 1.214x10⁻¹¹ = 0

Solving for X:

X: -3.48x10⁻⁶ → False answer, there is no negative concentrations

X: 3.484x10⁻⁶

As [OH⁻] = X; [OH⁻] = 3.484x10⁻⁶M

pOH = -log [OH⁻]; pOH = 5.46

As 14 = pOH + pH

pH = 8.54