Chemistry, 29.07.2019 21:30 manoli5874

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.

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Answer from: melinalange48

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

Answer from: DarinKing2005

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 %

Answer from: Quest

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh $pcl_{3}$ +δh $cl_{2}$ ] - [δh $pcl_{5}$ ]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

Answer from: Quest

What? i don’t understand

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

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