Problems to turn in for credit First Name Last Name: (Fall 2017) Clearly show yo

Question

Problems to turn in for credit First Name Last Name: (Fall 2017) Clearly show your work for full/partial credits. Report your answers to the correct number of significant figures Include units in your work and in your final answer when applicable. Make sure to draw a box around your final answer for questions involving calculations. Consider a 7.50 x 10- M solution of carbonic acid. (14 pts) a) 1. b) c) Determine the pH and the concentration of each species at equilibrium. what is the % dissociation of HOO'? what is the % dissociation of HCO

Explanation / Answer

Q1

carbonic acid has H2CO3 formula

a)

pH of each specie in equilbirium

First, assume the acid:

H2CO3

to be H2A, for simplicity, so it will ionize as follows:

H2A <-> H+ + HA-

HA- <-> H+ + A-2

where, H+ is the proton and HA- and A-2 are the conjugate bases, H2A is molecular acid

Ka1 = [H+][HA-]/[H2A]; by definition

Ka2 = [H+][A-2]/[HA-]

initially

[H+] = 0

[HA-] = 0

[H2A] = M;

the change

initially

[H+] = + x

[HA-] = + x

[HA2] = - x

in equilbrirum

[H+] = 0 + x

[A-] = 0 + x

[HA] = M - x

substitute in Ka

Ka1 = [H+][HA-]/[H2A]

Ka = x*x/(M-x)

x^2 + Kax - M*Ka = 0

if M = 0.10 M; then

x^2 + (4.5*10^-7)x - (7.50*10^-4)*(4.7*10^-7) = 0

solve for x

x =1.85*10^-5

substitute

[H+] = 0 + 1.85*10^-5= 1.85*10^-5

[HA-] = 0 + 1.85*10^-5 =1.85*10^-5

[H2A] = M - x = (7.5*10^-4 - 1.85*10^-5 ) = 0.0007315 M

Now...

Second ionization

HA- <-> H+ + A-2

initially

[H+] = x

[A-2] = 0

[HA-] = M-x

the change

[H+] = +y

[A-2] = + y

[HA-] = - y

in equilbrirum

[H+] = x+y

[A-2] = 0 + y

[HA-] = M - x - y

substitute in Ka

Ka2 = [H+][A-2]/[HA-]

Ka2 = (x+y)(y) / (M - x - y)

assume M-x >> M-x-y so (due to Ka2)

Ka2 = (x+y)(y) / (M - x)

y^2+ xy = Ka2(M-x)

y^2+ xy - Ka2(M-x) = 0

y^2+ (1.85*10^-5) *y - (4.7*10^-11)*(0.0007315 ) = 0

y = 1.858*10^-9

and we konw

[A-2] = + y

[CO3-2] = 1.858*10^-9

now,

[H+]total = 1.85*10^-5 + 1.858*10^-9= 0.0000185

[H2CO3] = 7.5*10^-4 - 1.85*10^-5 = 0.0007315

[HCO3-] = 1.85*10^-5- 1.858*10^-9 = 0.00001849

[CO3-2] = 1.858*10^-9

pH = -log(H) = -log(0.0000185) = 4.732

b)

% H2CO3 = 0.00001849 / (7.5*10^-4) * 100 = 2.465 %

c)

% HCO3- = (1.858*10^-9)/ (0.00001849) * 100 = 0.0100 %

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