Calculate the percent ionization of 0.0075 m butanoic acid (ka = 1.5 x 10-5) in a solution containing 0.085 m sodium butanoate.

The percent ionization of the solution is 0.017%.

The steps involved are;

Write the reaction equationSetup the ICE tablemake the necessary calculation

The equation of the reaction is;

            CH3CH2CH2COOH(aq)  ⇄ H^+(aq)  + CH3CH2CH2COO^-(aq)

I               0.0075                               0                0.085

C              -x                                        +x             + x

E          0.0075 - x                               x               0.085 + x

The Ka of the acid = 1.5 x 10^-5

Hence;

1.5 x 10^-5 = x(0.085 + x)/0.0075 - x

1.5 x 10^-5 (0.0075 - x ) =  x(0.085 + x)

1.1 x 10^-7 - 1.5 x 10^-5x = 0.085x + x^2

x^2 + 0.085x - 1.1 x 10^-7 = 0

x = 0.0000013 M

Percent ionization=  0.0000013 M/0.0075 × 100/1

= 0.017%

Learn more about percent ionization: link

Part 1: Calculate the percent ionization of 0.0075 m butanoic acid (Ka=1.5x10^-5) 
     
                 C4H8O2(aq) +H2O(l) → C4H7O2(aq) + H3O +
initial       0.0075                                   0                        0
change     -X                                        +X                      +X
final        0.0075-X                                  X                      X 
when Ka is relative smaller to the intial concentration of the acid so we can assume that 0.0075-X≈ 0.0075 by substitution in Ka formula:
Ka = [C4H7O2][H3O+] / [C4H8O2]
1.5x10^-5 = X*X / 0.0075

X^2 = 1.125x10^-7
X= 0.00034 m =3.4 x 10 ^-4 m
∴ [C4H7O2] = [H3O+] = 3.4X10^-4
 ∴ percent ionization = [H+ equlibrium]/[acid initial] *100
                                    = 3.4X10^-4/0.0075 * 100
                                    = 4.5 % 

part 2) calculate the percent ionization of 0.0075m butanoic acid in a solution containing 0.085m sodium butanoic?

                 C4H8O2(aq) + H2O(l) ↔ C4H7O2(aq) + H3O+
  initial        0.0075                                  0                 0.085
change         -X                                        +X                   +X
final           0.0075-X                                X                0.085+X
we can assume that 0.0075-X≈ 0.0075 & 0.085+X ≈ 0.085
∴Ka = (X*(0.085)) / (0.0075)
(1.5x10^-5)*0.0075 = 0.085X
∴X = 1.3x 10^-6
∴ percent ionization = (1.3x10^-6)/0.0075 * 100 = 0.017 %

 

Density and pressure

Bthats what im doing to and i figured it out it is b