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A 35.0-ml sample of 0.150 m acetic acid (ch3cooh) is titrated with 0.150 m naoh solution. calculate the ph after 17.5 ml of base have been added. the ka for the acetic acid equilibrium is 1.8 x 10-5. chegg

Respuesta :

superman1987
When the moles of CH3COOH = volume of CH3COOH * no.of moles of CH3COOH
moles of CH3COOH = 35ml * 0.15 m/1000 =0.00525 mol
moles of NaOH = volume of NaOH*no.of moles of NaOH
                           = 17.5 ml * 0.15/1000 = 0.002625
SO the reaction after add the NaOH:
                           CH3COOH(aq) +OH- (aq) ↔ CH3COO-(aq) +H2O(l)
initial                  0.00525                0                         0
change             - 0.002625         +0.002625     +0.002625
equilibrium      0.002625             0.002625        0.002625
When the total volume = 35ml _ 17.5ml = 52.5ml = 0.0525L
∴[CH3COOH] = 0.002625/0.0525 = 0.05m
and [CH3COO-]= 0.002625/0.0525= 0.05 m
when PKa = -㏒Ka
                 = -㏒1.8x10^-5 = 4.74
by substitution in the following formula:
PH = Pka + ㏒[CH3COO-]/[CH3COOH]
      = 4.74 + ㏒(0.05/0.05) = 4.74
∴PH = 4.74
BladeRunner212

5 - log 1.8 or 4.74.

Further explanation

Given:

  • A 35.0-ml sample of 0.150 M acetic acid (CH₃COOH) is titrated with 0.150 M NaOH solution.  
  • After 17.5 ml of base have been added.
  • The Ka for the acetic acid equilibrium is 1.8 x 10⁻⁵.

Question:

Calculate the pH of this buffer.

The Process:

Step-1

Let us prepare the mole number of the two reagents.

[tex]\boxed{ \ n = MV \ }[/tex]

Moles of CH₃COOH = [tex]\boxed{ 0.150 \ \frac{mol}{L} \times 35.0 \ ml = 5.25 \ mmol \ }[/tex]

Moles of NaOH = [tex]\boxed{ 0.150 \ \frac{mol}{L} \times 17.5 \ ml = 2.625 \ mmol \ }[/tex]

Step-2

Let use the ICE table.

Balanced reaction:

                   [tex]\boxed{ \ CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O \ }[/tex]

Initial:                5.25           2.625                -                  -

Change:           -2.625        -2.625          +2.625        +2.625

Equlibrium:      2.625             -                 2.625          2.625

  • NaOH as a strong base acts as a limiting reagent.
  • The remaining CH₃COOH as a weak acid and CH₃COONa salt form an acidic buffer system.
  • The CH₃COONa salt has valence = 1 according to the number of CH₃⁻ ions as a weak part, i.e., [tex]\boxed{ \ CH_3COONa \rightleftharpoons CH_3COO^- + Na^+ \ }[/tex]
  • CH₃COOH and CH₃COO⁻ are conjugate acid-base pairs.

Step-3

Let us prepare pKa and concentrations of CH₃COOH and CH₃COO⁻.

[tex]pK_a = -log(Ka)[/tex]

[tex]\boxed{ \ pK_a = -log[1.8 \cdot 10^{-5}] = 5 - log \ 1.8 \ }[/tex]

Total volume system = 35.0 ml + 17.5 ml = 52.5 ml.

[tex]\boxed{ \ [CH_3COOH] = \frac{2.625 \ mmol}{52.5 \ ml} = 0.05 \ M \ }[/tex]

[tex]\boxed{ \ [CH_3COO^-] = \frac{2.625 \ mmol}{52.5 \ ml} = 0.05 \ M \ }[/tex]

Step-4

And now it's time we calculate the pH value of this buffer system.

[tex]\boxed{ \ pH = pK_a + log\frac{[CH_3COO^-]}{[CH_3COOH]} \ }[/tex]

[tex]\boxed{ \ pH = 5 - log \ 1.8 + log\frac{[0.05]}{[0.05]} \ }[/tex]

[tex]\boxed{ \ pH = 5 - log \ 1.8 + log \ 1 \ }[/tex]

[tex]\boxed{ \ pH = 5 - log \ 1.8 + 0 \ }[/tex]

Thus, we get the pH value of this buffer system which is equal to 5 - log 1.8 or 4.74.

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