Question

I'm not sure how to solve this! Any help is appreciated!  

1.Draw the pH titration curve for the titration of 35 mL of 0.150 M acetic acid with 0.200 m Include the pH values and NaOH volume requested below on your pH titration curve. Al curve the species that dictates the pH at the requested pH values. For acetic acid, Ka = 1.885 in your pH titration curve. Also, put on the 1) the initial pH 2) the pH after 15.0 mL of NaOH have been added 3) the volume of NaOH at the equivalence point 4) the pH at the equivalence point 5) the pH after 50.0 mL of NaOH have been added 1) AA + NaOH + H₂O + A PH mh of .2 NaOH

Answer 1

1) the initial pH

Initially there is only acetic acid present in the solution, hence pH of solution will be due to dissociation of acetic acid. Consider a dissociation of acetic acid in water.

CH₃COOH _(aq) + H₂O _(l) CH₃COO ^-_(aq) + H₃O ⁺_(aq)  

For above reaction, dissociation constant is Ka = [CH₃COO ^- ] [H₃O ⁺ ] / [CH₃COOH] = 1.8 $\times$ 10 ^-05

Consider X moles of acetic acid dissociated at equilibrium, then we can write

K a = (X) $\times$ ( X) / 0.150 - X = 1.8 $\times$ 10 ^-05

Acetic acid is a weak acid. Hence, we can assume X is very small as compared to 0.150. Hence we can write 0.150 - X 0.150.

X ² / 0.150 = 1.8 $\times$ 10 ^-05

  X ² = 0.150 ( 1.8 $\times$ 10 ^-05 )

  X ² = 2.7 $\times$ 10 ^-06

X = 1.64 $\times$ 10 ^-03 M = [CH₃COO ^- ] =[H₃O ⁺ ]

We have, pH = -log [H₃O ⁺ ] = - log 1.64 $\times$ 10 ^-03 = 2.78

2) pH of solution after addition of 15.0 ml NaOH

Consider a reaction , CH₃COOH _(aq) + NaOH _(aq)  CH₃COONa _(aq) + H₂O (l)

mmol of CH₃COOH = concentration x volume = 0.150 $\times$ 35 =5.25 mmol

mmol of NaOH = 0.200 $\times$ 15.0 = 3.00 mmol

Excess mmol of CH₃COOH = initial mmol - mmol after reaction with NaOH = 5.25 - 3.00=2.25

mmol of CH₃COONa produced = mmol of NaOH added to solution = 3.00

Volume of solution =35.0 ml +15.0 ml = 50.0 ml

[CH₃COOH]= No of moles / Volume of solution in L

[CH₃COOH] = [2.25 /1000] / [50.0 /1000]

[CH₃COOH ] = 0.045 M

[CH₃COONa]= [3.00 /1000] / [50.0 /1000]

[CH₃COONa]= 0.0600 M

At this stage, solution contain weak acid (acetic acid) and its salt (sodium acetate) .Hence, solution acts as a buffer solution and it's pH is calculated by using Henderson's equation

pH = pKa + log [ salt] / [ Acid]

pH = - log Ka + log [CH₃COONa] / [CH₃COOH]

pH = -log (1.8 x 10 ^-05) + log 0.0600 /0.045

pH = 4.74 + 0.125

pH = 4.86

3) the volume of NaOH at the equivalence point

Consider a reaction , CH₃COOH _(aq) + NaOH _(aq)  CH₃COONa _(aq) + H₂O (l)

From reaction, 1 mol CH₃COOH $\equiv$ 1 mol NaOH

We have relation, M _acid V _acid = M _base V _base

   V _base = M _acid V _acid / M _base

   V _base =0.150 M 35.0 ml / 0.200 M

   V _base = 26.25 ml

Volume of NaOH at equivalence point = 26.25 ml

4) the pH at the equivalence point

At equivalence point, all acetic acid is consumed by added NaOH. pH of solution will be due to dissociation of CH₃COONa in water.

CH₃COO^- (aq) + H₂O _(l) CH₃COOH _(aq) + OH ^-_(aq)

Kb = [CH₃COOH] [OH ^- ] / [CH₃COO^- ]= Kw / Ka = 10 ^-14 / 1.8 $\times$10 ^-05 = 5.56 $\times$ 10 ^-10

mmol of CH₃COONa produced = mmol of NaOH added to the solution =0.200 $\times$ 26.25  =5.25 mmol

Volume of the solution = volume of acetic acid + volume of NaOH = 35.0 ml +26.25 ml  = 61.25 ml

[CH₃COONa]= [5.25/1000] / [61.25 /1000] = 0.0857 M

Consider X moles of sodium acetate dissociated at equilibrium, then we can write

Kb = (X ) ( X) / 0.0857 - X = 5.56 $\times$ 10 ^-10

Here X is very small as compared to 0.857. Hence we can write 0.0857 - X 0.0857.

X ² = 0.0857 ( 5.56 $\times$ 10 ^-10 )  

X ² = 4.76$\times$ 10 ^-11

X = 6.90 $\times$10 ^-06 M = [ OH ^- ]

We know that , pOH = -log [ OH ^- ] = - log 6.90 $\times$10 ^-06 = 5.16

We have relation, pH + pOH = 14

Therefore, pH = 14 - pOH = 14 - 5.16 = 8.84

5) pH after 50.0 ml NaOH added

After equivalence point, pH of solution will be due to NaOH.

mmol of NaOH =0.200 $\times$ 50.0 = 10.0 mmol

mmol of CH₃COOH =0.150 $\times$ 35.0 = 5.25 mmol

mmol of excess NaOH = 10.0 - 5.25 = 4.75 mmol

Volume of solution = volume of acid + volume of base = 35.0 + 50.0 = 85.0 ml

[NaOH ] = [ 4.75 /1000] / [ 85.0 /1000] = 0.05588 M

We know that , pOH = -log [ OH ^- ] = - log 0.0558 = 1.31

We have relation, pH + pOH = 14

Therefore, pH = 14 - pOH = 14 - 1.31 = 12.69

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