Chem. 127

Homework #3

Distributed Wed 10/13

Due Monday Oct 18

 

 

  1. Calculate the pH of 0.025 M Ca(OH)2
  2. Calculate the pH of 0.01M hydrocyanic acid (HCN) (Ka=4.9*10-10 ) and percent dissociation.
  3. Calculate the pH and concentration of all species presented in 0.4M phosphoric acid (H3PO4). Use table15.3 p.631 in your textbook
  4. Using values of Ka in Appendix C, calculate values of Kb for each of the following ions:

            a)Acetate ion CH3COO-

            b) Iodide ion I-

c)      Sulfate ion SO42-

       5.Predict whether the following salt solutions are neutral or acidic or basic and calculate pH?

A)    0.3M NiCl2   B) 0.05M NaNO2 (Use Appendix C)

 

  1. p.657 15.97

 

 

Solutions to the problems

1. pH of 0.025 M Ca(OH)2

this is a strong base, 100% dissociated

two hydroxyl per Ca(OH)2 , thus, [OH-] =0.050 M

            pOH = -log10( 0.050) = 1.30

            pH = 14.00 – 1.30 = 12.70

0.025 is two sig figures, so we should have two degits behind decmal on pH, pOH

 

also valid… [H+] = 1.x 10-14 / 0.050 =  2.0 x 10-13

pH = -log10(2.0 x 10-13) = 12.70

 

2. pH of 0.01M hydrocyanic acid (HCN) (Ka=4.9*10-10 )

                        x= [H3O+]   x= [CN-]   0.010 –x = [HCN]

                        Ka is quite small, I expect x << 0.010

 

                        Approx Ka = (x)*(x) / (0.010 –x) = = (x)*(x) / (0.010 __)

                        X = (4.9 x 10-10 x 0.010)1/2 = 2.2 x 10-6

                                    Approximation is well justified

                        PH= -log(2.2 x 10-6) = 5.56

 

            fraction dissociation   [H2O+] / [HCN, original] = 2.2 x 10-6 / (0.010 ) = 2.2 x 10-4

            percent is 0.022 %

 

3. pH and concentration of all species presented in 0.4M phosphoric acid (H3PO4).

 

            First, the [H3O+] and pH fixed by strongest acid in a mixture

            For a triprotic acid, this means Ka1 fixes pH

                        We will make several approximations

                        Perhaps, [H3O+] << 0.40 (usual weak acid approximation)

                        [H3O+] form step 2 and 3 is negligible

                        [H2PO4-] << [HPO4-2]

                        [PO43-] << [HPO42-]

            (like with calculators, I’ll start using 7.5 E-3 instead of typing 7.5 x 10-3 with exponents)

            ok,  Ka1 =  7.5 E-3   = [x] [x] / (0.40 –x) for a weak acid

                        x = (approx) sqrt( 0.4 x  7.5 E-3 ) = 0.055

                                    that’s a marginal approximation, I need to solve as a quadratic here

 

                                    another trick is an iterative solution

                                    approximating 0.40-x with 0.40 is a poor approximation

                                    approximating it with 0.40 – 0.055 is a better guess now

                                                x = (now) sqrt( 0.345 x 7.5E-3) = 0.0509

                                                notice I’m really making a small error now

                                                x = (now) sqrt( 0.341 x 7.5E-3) = 0.051

                                                within 2 significant figures, no error

                                    x = [H3O+] = 0.051 M

                                    [H2PO4-]  also = 0.051

pH= -log(.051) = 1.29

            now look at step 2

                        [H2PO4-] + H2O ŕ [H3O+]  +  [HPO4-2]

                        (0.051 – x) = [H2PO4-]

                        (0.051 + x) = [H3O+]  

                        (x) =  [HPO4-2]

                                    Ka2 = 6.2 E-8 = (0.051+x)(x)/ (0.051 –x)

                                    Almost  6.2e-8 = (0.051)(x)/ (0.051)

                                                Or  x = 6.2 x 10-8   and the approximations are fine

 

            Finally, equil #3

                        [HPO4-2]  + H2O ŕ[H3O+]  + [PO43-]

                        Ka3 = 4.8 e –13

                         [PO43-] = x

[HPO4-2]  = 6.2 e-8  -x

[H3O+]  = 0.051 +x

                                    Ka = (almost)  = (0.051) (x) / (6.2e-8)

                                    X = (4.8e-13) x (6.2e-8)/ (0.051)

                                    [PO43-] = x = 5.8 e –19

                                    compare with Avogadro’s numer—it’s only about 10,000 ions in a liter

 

            You could also use Ka1 x Ka2 x Ka3 = [H3O+]3  (x) / [H3PO4]

 

  1. #3. calculate values of Kb for each of the following ions:

            a)Acetate ion CH3COO

Ka 9acetic acid) 1.8e-5

Kb = 1.0 E-14 / Ka =

            b) Iodide ion I-

                              conjugate acid would be HI

tables don’t list a Ka value

if strong acid, then no Kb value (or formally Kb = 0)

Ka (HI) =

d)      Sulfate ion SO42-

A)    The conjugate acid is HSO4-  (not H2SO4)

B)     Use Ka2 for sulfuric acid = 1.2 x 10-2

C)    Kb = 1.0 x10-14 / 1.0 x 10-2 = 1.0 x 10-12

 

7. Solutions (without numerical results)

            Table C2 lists Ka 2.5 x 10-11 for Ni2+     

                        Ni(H2O)62+ --ŕ Ni(OH)(H2O)52+  + H3O+

                        A weak acid , Ka given and [acid] = 0.30 M

                                    Cl- ion is a spectator, has no acid/base role

 

            Table C1 lists HNO2 as a weak acid, Ka = 4.5 x 10-4

                        As in #6 Kb for HNO2 = 1.0 x10-14/Ka

                        Weak base in water, conc [base] = 0.050

                                    Na+ is a spectator ion, no acid base role