1] When bitten by a red ant, the stinging sensation is caused by formic acid, th
ID: 897615 • Letter: 1
Question
1] When bitten by a red ant, the stinging sensation is caused by formic acid, the simplest of the organic weak acids. The dissociation reaction is:
A 0.010 M solution of formic acid has a pH = 2.87.
Complete the following table and use the data to determine the Ka for the weak acid. NOTE: SMALL VALUES CAN ENTERED EITHER AS THE DECIMAL (e.g., 0.0000232) OR EXPONENTIAL (2.32E-5). Keep at least 2 digits after the zeros and do the subtraction on the final line for the acid
Ka = ___________
b. What is the % ionization of this solution?
___________%
2]Benzoic acid is C6H5CO2H and has a Ka of 6.8 x 10-5. What is:
a.the pH of a 0.500 M solution of benzoic acid?
b. the pKb of the benzoate ion (the conjugate base>
pH = __________
pKb = __________
3]A 5-6% solution of acetic acid (CH3CO2H) is sold as vinegar. This weak organic acid hs a Ka of 1.8 x 10-5.
a. Determine the equilibrium concentrations of all the species in a 0.19 M acetic acid solution.
The concentrations of the species are: H3O1+ = _______ M
CH3CO21- = ________ M
CH3CO2H = ________M
b. What is the pKa of this acid?
pKa = _________
c. What is the pH of this solution?
pH = _________
d. What is the %ionization of this solution?
%ionization = __________
4] The percent ionization refers to the percentage of hydronium ion resutling from the dissociation of a weak acid. If the Kaof acetic acid (CH3COOH) is 1.8 x 10-5, what is the percent ionization of:
a. a 0.09 M solution?
b. a 0.0001 M solution?
% ionization of solution a = _______________
% ionization of solution b = _______________
Explanation / Answer
Solution :-
Q1) reaction equation
HCO2H (aq)+ H2O (l) -----> HCO2^- (aq) + H3O^+ (aq)
0.01M solutioin has pH =2.87
From the given pH lets calculate the [H3O+]
[H3O+] = antilog [-pH]
[H3O+] = antilog [-2.87]
= 0.001349 M
Following is the values in the table
HCO2H (aq)+ H2O (l) -----> HCO2^- (aq) + H3O^+ (aq)
Initial 0.01 M 0 0
Change -0.001349 0.001349 0.001349
Equilibrium 0.01-0.001349 0.001349 0.001349
=0.008651
Now lets calculate the ka
Ka= [H3O+] [HCO2^-] /[HCO2H]
Ka = [0.001349][0.001349] /[0.008651]
Ka= 2.10*10^-4
b) % ionization = (x/initial concentration) *100%
= (0.001349/0.01)*100%
= 13.5 %
Q2) 2]Benzoic acid is C6H5CO2H and has a Ka of 6.8 x 10-5. What is:
a.the pH of a 0.500 M solution of benzoic acid?
Solution :-
ka = [H3O+] [C6H5CO2-] /[C6H5CO2H]
ka= [x][x]/ [C6H5CO2H-x]
since ka is very small therefore we can neglect the x from the denominator
ka= [x][x]/ [C6H5CO2H-]
6.8*10^-5 = x^2 / [0.500]
6.8*10^-5 * 0.500 = x^2
3.4*10^-5 = x^2
Taking square root of both sides we get
5.83*10^-3 = x =[H3O+]
pH= -log [H3O+]
pH= -log [5.83*10^-3]
pH= 2.23
b. the pKb of the benzoate ion
using the given ka lets calculate the pka
pka = -log ka
pka = -log 6.8*10^-5
pka = 4.17
now pka + pkb= 14
therefore
pkb = 14 – pka
pkb = 14 -4.17
pkb = 9.83
3]A 5-6% solution of acetic acid (CH3CO2H) is sold as vinegar. This weak organic acid hs a Ka of 1.8 x 10-5.
a. Determine the equilibrium concentrations of all the species in a 0.19 M acetic acid solution.
Ka= [H3O+] [CH3COO-]/[CH3COOH]
Ka= [x][x]/ [CH3COOH]
1.8*10^-5 = x^2 / 0.19 M
1.8*10^-5 * 0.19 = x^2
3.42*10^-6 = x^2
Taking square root of the both sides we get
1.85*10^-3 =x [H3O+]
Therefore the equilibrium concentrations are
[H3O+] = 1.85*10^-3 M
The concentrations of the species are: H3O1+ = 1.85*10^-5 M
CH3CO21- = 1.85*10^-3M
CH3CO2H = 1.88*10^-1M
b. What is the pKa of this acid?
pKa = -log ka
Pka = -log 1.8*10^-5
Pka = 4.74
c. What is the pH of this solution?
pH = -log [H3O+]
= -log [1.85*10^-3]
= 2.73
d. What is the %ionization of this solution?
%ionization = (x/initial conc) *100%
= (0.00185 / 0.19)*100%
= 0.97 %
4] The percent ionization refers to the percentage of hydronium ion resutling from the dissociation of a weak acid. If the Kaof acetic acid (CH3COOH) is 1.8 x 10-5, what is the percent ionization of:
a. a 0.09 M solution?
b. a 0.0001 M solution?
Solution :-
lets first calculate the concentration of the H3O+ for the each solution
a. Ka= [H3O+] [CH3COO-]/[CH3COOH]
Ka= [x][x]/ [CH3COOH]
1.8*10^-5 = x^2 / 0.09 M
1.8*10^-5 * 0.09 = x^2
1.62*10^-6 = x^2
Taking square root of both sides we get
1.27*10^-3 = x
Therefore the percent ionization of the a is as follows
%ionization = (x/initial conc) *100%
= (0.00127 / 0.09)*100%
= 1.41 %
Calculating for the b
Ka= [H3O+] [CH3COO-]/[CH3COOH]
Ka= [x][x]/ [CH3COOH]
1.8*10^-5 = x^2 / 0.0001M
1.8*10^-5 * 0.0001 = x^2
1.8*10^-9 = x^2
Taking square root of both sides we get
4.24*10^-5 = x
Therefore the percent ionization of the a is as follows
%ionization = (x/initial conc) *100%
= (4.24*10^-5 / 0.0001)*100%
= 42.4 %
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