Question

2. Calculate the pH of a 0.230 M calcium acetate solution if the Ka for acetic acid is 1.8x10-5 .

3. Calculate the pH of a 0.225 M solution of H2CO3. Ka1 = 4.3x10^-7 ; Ka2 = 5.6x10^-11

4. Calculate the pH of a 0.0080 M solution of sulfuric acid. Ka2 = 1.2x10^-2

Answer 1

ANSWER;-

1)-calcium acetate is a weak base, a conjugate base of acetic acid, so:

Ka × Kb = K_W that is

K_b^acetate = K_w / K_a^{acetic acid} =

= 10^-14 / 1.8 x 10^-5

= 5.376 × 10^-10

In this case c_{weak base} >>> K_b, that is the equation to use is:

[OH^– ] = $\sqrt[]{}$
KCweakbase
=
5.376X 10-10 X0.230

[OH^– ] = 1.2365 X 10^-5

pOH = -log[OH^- ]

= -log(1.2365 × 10^-5) = +4.907

pH = 14.00 – pH = 14.00 – 4.907 =9.093

2)- For H₂CO₃, Ka1 = 4.3 x 10^–7 and Ka2 = 5.6 x 10^–11.

1st ionization equation: H₂CO₃ (aq) + H₂O (l) ⇄ HCO₃ – (aq) + H₃O+ (aq)

2nd ionization equation: HCO₃^– (aq) + H₂O (l) ⇄ CO3 ^2– (aq) + H₃O+ (aq)

because K_a1 > K_a2 & K_a1 >> K_w, the primary source of H₃O+ in the solution will be the 1st ionization step for H₂CO₃

H₂CO₃ (aq) + H₂O (l) ⇄ HCO₃ – (aq) + H₃O+ (aq)

initial 0.040 M 0 0

∆   ∆ + x   + x

equil (0.225–x)M   x M x M

Ka = [H+] [HCO3-] / [CO2]

=4.3x10^-7 = x² / (0.225-x)

Ka   = 3.11 X 10^-4

  [H+] =3.11 X 10^-4

pH = -log[H+] = -log(3.11 X 10^-4 ) = +3.507

pH = +3.507

Just in case you weren't sure about that....

HCO3- <==> H+ + CO3^3- ..... Ka = 5.6x10^-11

3.11 X 10^-4....3.11 X 10^-4 ..0 ...... initial

-x ................. +x .........+x .............. change

(3.11 X 10^-4-x) (3.11 X 10^-4+x ) x ..... equilibrium

Ka = [H+] [CO3^2-] / [HCO^3-]

5.6x10^-11 = (3.11 X 10^-4+x)(x) / (3.11 X 10^-4-x)

x IS VERY SMALL SO IT WILL ZERO IN + AND - TERM

THEN CALCULATE X = 5.6 X 10^-11

3-

HSO4^-+ H2O $\rightleftharpoons$ H3O⁺ + SO4^2-  

intial 8.0 x 10^-3 8.0 x 10^-3 0

change -x +x +x

equill (8.0 x 10^-3 -x) (8.0 x 10^-3 +x) +x

Ka = [H3O⁺] [SO4^2-] / [HSO4^-]

0.0120 = [X]² / (0.0080 -X)
So [X]² = 0.0120*( 0.0080 - X)
X= 9.8 x 10^-3

pH = -log [H+]

pH = -log (9.8 x 10^-3)

pH = 2.00