Butanoic acid , HC 4 H 7 O 2 , is a weak acid . The acid dissociation constant ,

Question

Butanoic acid , HC4H7O2 , is a weak acid . The acid dissociation constant , Ka, is (1.5*10^-5 ) .

a) Calculate the [H+] of a 0.42-molar solution of HC4H7O2.

b) Write the balanced net ionic equation for the reaction that occurs NaC4H7O2 is dissolved in water and calculate the numerical value of the equilibrium constant for this reaction.

c) Calculate the pH of a solution made by combining 40.0 milliliters of 0.42-molar HC4H7O2 and 5.00 milliliters of 1.68-molar NaOH.

d) How many grams of solid NaC4H7O2 must be added to 50.0 milliliters of 0.800-molar HC4H7O2 to obtain a buffer solution that has a pH of 4.82? Assume that the addition of the solid NaC4H7O2 results in a negligible change in volume.

e) Household bleach is made by dissolving chlorine gas in water , as represented below.

Cl2(g) + H2O -> H+ Cl- + HOCl(aq)

- Note that this reaction is not an equilibrium reaction. It does go to completion .

Calculate the pH of such a solution if the concentration of HOCl in the above solution is 0.005 molar. (ignore any equilibrium reactions that result from the partial dissociation of HOCl).   

(PLEASE SHOW ALL YOUR WORK )

THANK YOU :) !!!

Explanation / Answer

a.

HC4H7O2 + H2O <=> C4H7O2^- +     H3O+

   0.42               /                     0                      0

     - x                                    +x                       +x

    0.42-x                                 x                        x

In 0.42-x, x is insignificant

Ka = [C4H7O2^-][H3O+]/[HC4H7O2]

      = x^2/0.42

      = 1.5 x 10^-5

x = 0.0025 mol/L = [H+]

[H+] = 0.0025 mol/L

b.

C4H7O2^-   +    H2O   <=>     HC4H7O2    + OH-

Na+ is not shown as it is a spectator ion.

The name of the anion is butanoate, and is a conjugate base of butanoic acid.

The reaction is the hydrolysis of the conjugate base, thus its equilibrium constant can be calculated by

Kb = Kw/Ka

       = 1 x 10^-14 / 1.5 x 10^-5

       = 6.7 x 10^-10

c.

There is

0.42 M x 0.040 L = 0.0168 mol HC4H7O2

1.68 M x 0.005 L = 0.0084 mol NaOH

When these two react,

HC4H7O2     +     NaOH       -->     NaC4H7O2     + H2O

   0.0168               0.0084                      0                       /

-0.0084               -0.0084                +0.0084              

   0.0084                       0                     0.0084

There is 0.0084 mol of both HC4H7O2 and C4H7O2 in the resulting solution.

To calculate pH, you would use the henderson-hasselbalch equation which is

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

In this case, since mol base = mol acid,

[base] = [acid]

[base]/[acid] = 1

log 1 = 0

Thus

pH = pKa        This is always the case when there is equal amount of weak acid and its conjugate base.

pKa = -log Ka = log 1.5 x 10^-5

        = 4.82

d.

Since we know pKa = 4.82, and we want buffer with pH of 4.82,

we know that we need equal amount of acid and conjugate base.

There is weak acid in the solution and we are adding the conjugate base

we just need to make sure we add the same amount

There is

0.800 M x 0.050 L = 0.040 mol acid

We need 0.040 mol of NaC4H7O2

molar mass of NaC4H7O2 = 110.09 g/mol

0.040 mol x (110.09 g/mol) = 4.40 g

We need to add 4.40 g NaC4H7O2

e.

This is a very simple question if we ignore both reaction equilibrium and HOCl equilibrium

Cl2 + H2O --> H+    + Cl- + HOCl

H+, Cl-, and HOCl are formed in 1:1:1 mol ratio.

This means that if [HOCl] = 0.005 M, [H+] = 0.005 M also.

pH = -log [H+] = -log (0.005)

      = 2.30

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