A 50.0 ml sample of 0.3 M acetic acid, HC2H3O2, is titrated with 0.1500 M NaOH.

Question

A 50.0 ml sample of 0.3 M acetic acid, HC2H3O2, is titrated with 0.1500 M NaOH. Ka(Acetic Acid) = 1.8 x 10^-5 a) Write a balanced net-ionic equation for the reaction of the above titration.....b) What is the pH before any NaOH is added?.........c) What is the pH half way to the equivalence point?.......d) How many mls of NaOH are required to reach the equivalence point?.....e) How many moles of acetate are formed at the equivalence point and what is the concentration of acetate C2H3O2 at the equivalence point?.......f) What is the pH at the equivalence point?

Please show all work, thank you:)

Explanation / Answer

A 50.0 ml sample of 0.3 M acetic acid, HC2H3O2, is titrated with 0.1500 M NaOH. Ka(Acetic Acid) = 1.8 x 10^-5

a) Write a balanced net-ionic equation for the reaction of the above titration.....

solution :-

Balanced reaction equation

CH3COOH + NaOH ------- > CH3COONa + H2O

b) What is the pH before any NaOH is added?.........

Solution :-

before adding any NaOH

CH3COOH + H2O ------ > CH3COO- + H3O^+

0.3 M                                       0                  0

-x                                              +x                +x

0.3-x                                            x               x

Ka = [H3O+][CH3COO-]/[CH3COOH]

1.8*10^-5 = [x][x]/[0.3-x]

Since ka is small we ccan neglect the x from denominator

1.8*10^-5 = [x][x]/[0.3]

1.8*10^-5 * 0.3 = x^2

5.4*10^-6 = x^2

Taking square root of both sides we get

2.32*10^-3 =x

pH= -log [H3O+]

pH= -log [2.32*10^-3]

pH= 2.63

c) What is the pH half way to the equivalence point?.......

solution :-

At the half way equivalence point half of the acid is converted to its conjugate base so the ratio of the conjugate ase to acid is 1

So the pH= pka

Pka = - log ka

       = -log 1.8*10^-5

       = 4.74

So pH= 4.74

d) How many mls of NaOH are required to reach the equivalence point?.....

Solution :- at the equivalence point moles of acid and moles of base are same

so the volume of NaOH needed = volume of acid * moalrity if acid / molarity of NaOH

                                                           = 50.0 ml * 0.30 M / 0.1500 M

                                                          = 100 ml

so the volume of NaOH needed = 100 ml

e) How many moles of acetate are formed at the equivalence point and what is the concentration of acetate C2H3O2 at the equivalence point?.......

solution :-

moles of acetate = moles of acetic acid

moles of acetic acid = molarity * volume in liter

                                      = 0.30 mol per L * 0.050 L

                                      = 0.015 mol acetic acid

So the moles of acetate formed at the equivalence point = 0.015 mol acetate

f) What is the pH at the equivalence point?

solution :- Lets find the molarity of the acetate at the total volume

Total volume = 50 ml + 100 ml = 150 ml

Molarity of the acetate = 0.015 mol / 0.150 L = 0.10 M

CH3COO-   + H2O ----- > CH3COOH + OH-

0.10                                          0                0

-x                                              +x               +x

0.10-x                                      x                x

kb = [x][x]/[0.10-x]

kb of acetate = 5.55*10^-10

5.55*10^-10 = [x][x]/[0.10-x]

Since kb is small we can neglect the x from denomintaotr then we get

5.55*10^-10 = [x][x]/[0.10-]

5.55*10^-10 * 0.10 = x^2

5.55*10^-11 = x^2

Taking square root of both sides we get

7.45*10^-6 =x

Now lets find pOH

pOH = -log [OH-]

       = -log [7.45*10^-6]

      = 5.13

pH + pOH = 14

pH= 14 – pOH

pH= 14 – 5.13

pH= 8.87

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