Chem. 127

Homework #3

Distributed Wed 10/13

Due Monday Oct 18

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

a)Acetate ion CH_{3}COO^{-}

b) Iodide ion I^{-}

c)
Sulfate ion SO_{4}^{2-}

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

A)
0.3M NiCl_{2 } B) 0.05M NaNO_{2 }(Use Appendix C)

- 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 [H_{2}O+] / [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 (H_{3}PO_{4}).

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,
[H_{3}O+] << 0.40 (usual weak acid approximation)

[H_{3}O+]
form step 2 and 3 is negligible

[H_{2}PO_{4}^{-}]
<< [HPO_{4}^{-2}]

[PO_{4}^{3-}]
<< [HPO_{4}^{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
= [H_{3}O+] = 0.051 M

[H_{2}PO_{4}^{-}]
also = 0.051

pH= -log(.051) = 1.29

now look at step 2

[H_{2}PO_{4}^{-}]
+ H2O ŕ
[H_{3}O+] + [HPO_{4}^{-2}]

(0.051
– x) = [H_{2}PO_{4}^{-}]

(0.051
+ x) = [H_{3}O^{+}]

(x)
= [HPO_{4}^{-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

[HPO_{4}^{-2}] + H2O ŕ[H_{3}O+] + [PO_{4}^{3-}]

Ka3 = 4.8 e –13

[PO_{4}^{3-}] = x

[HPO_{4}^{-2}] = 6.2 e-8
-x

[H_{3}O+] = 0.051 +x

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

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

[PO_{4}^{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 = [H_{3}O+]^{3} (x) / [H_{3}PO_{4}]

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

a)Acetate ion CH_{3}COO

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 SO_{4}^{2-}

A)
The conjugate acid is HSO_{4}- (not H_{2}SO_{4})

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 Ni^{2+}

Ni(H_{2}O)_{6}^{2+}
--ŕ
Ni(OH)(H_{2}O)_{5}^{2+} + H_{3}O^{+}

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