Question

Determine the pH (to two decimal places) of the following solutions. If the 5% approximation is valid, use the assumption to compute pH.

8.72×10^-3 M sodium hydrogen sulfate

pKa hydrogen sulfate ion is 1.92

8.73×10^-3 M trifluoroacetic acid

pKa trifluoroacetic acid is 0.20

8.20×10^-3 M ethylamine

pKa ethylammonium ion is 10.87

2.51×10^-1 M sodium nitrite

pKa nitrous acid is 3.37

Answer 1

1) 8.72×10^-3 M sodium hydrogen sulfate

pKa hydrogen sulfate ion is 1.92

solution :- from the pka we can calculate the ka

pka= -log ka

ka= antilog [-pka]

    = antilog [-1.92]

    = 0.0120

HSO4 - + H2O ----- > H3O+ + SO4^2-

8.72*10^-3                     0                0

-x                                       +x         +x

8.72*10^-3 -x                x               x

Ka= [H3O+] [SO4^2-]/[HSO4-]

0.0120 = [x][x]/[8.72*10^-3-x]

0.0120 * [8.72*10^-3-x]= x^2

Solving for the x we get

x= 0.0059 M = H3O+

using this we can calculate the pH

pH= -log [H3O+]

   = -log [0.0059]

   = 2.23

2) 8.73×10^-3 M trifluoroacetic acid

pKa trifluoroacetic acid is 0.20

Solution :-

pka= -log ka

ka= antilog [-pka]

    = antilog [-0.2]

    = 0.631

Lets assume the acid is as HA

HA + H2O ------ > H3O+ + A-

8.73*10^-3              0            0

-x                             +x         +x

8.73*10^-3 -x           x           x

Ka= [H3O+] [A-]/[HA]

0.631 = [x][x]/[8.73*10^-3-x]

0.631 * [8.73*10^-3-x]= x^2

Solving for the x we get

x= 0.0086 M = H3O+

using this we can calculate the pH

pH= -log [H3O+]

   = -log [0.0086]

   = 2.07

3) 8.20×10^-3 M ethylamine

pKa ethylammonium ion is 10.87

Solution :-

Pka = 10.87

From pka lets find pkb

Pkb = 14 – 10.87 = 3.13

Kb = antilog [-3.13]

Kb = 0.000741

Now lets write the equation

C2H5NH2 + H2O ------- > C2H5NH3+ + OH-

8.20*10^-3                              0                       0

-x                                             +x                    +x

8.20*10^-3 -x                           x                       x

Kb= [C2H5NH3+ ] [OH-] /[ C2H5NH2]

0.000741 = [x][x]/[8.20*10^-3-x]

0.000741 * [8.20*10^-3-x] = x^2

Solving for x we get

X= 0.00212 = [OH-]

Now lets find pOH

pOH= -log [OH-]

pOH = -log [0.00212]

         = 2.67

pH + pOH = 14

pH= 14 – pOH

   = 14 – 2.67

   = 11.33

4) 2.51×10^-1 M sodium nitrite NaNO2

pKa nitrous acid is 3.37

Solution :-

Pkb = 14 – 3.37 = 10.63

Kb = antilog [-10.63] = 2.34*106-11

NO2- + H2O ----- > HNO2 + OH-

0.251                             0           0

-x                                   +x          +x

0.251-x                        x             x

Kb = [HNO2][OH-]/[NO2-]

2.34*10^-11 = [x][x]/[0.251-x]

Since kb is very small therefore we can neglect the x from the denominator

So we get

2.34*10^-11 = [x][x]/[0.251]

2.34*10^-11 * 0.251 = x^2

5.87*10^-12 =x^2

Taking square root of both sides

2.42*10^-6 = x

pOH = -log [OH-]

pOH = -log [2.42*10^-6 ]

pOH = 5.62

pH= 14 – pOH

pH= 14 – 5.62

pH = 8.38