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 SO₄^2-

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)

5. 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]

[PO₄^3-] << [HPO₄^2-]

(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 à [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 à[H₃O+] + [PO₄^3-]

Ka3 = 4.8 e –13

[PO₄^3-] = 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)

[PO₄^3-] = 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]

6. #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 SO₄^2-

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⁺ --à 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