Acid/Base Equilibrium - Weak Acids/Bases 1. Calculate the [HClO] in a solution i
ID: 896837 • Letter: A
Question
Acid/Base Equilibrium - Weak Acids/Bases
1. Calculate the [HClO] in a solution if it has a pH 5.35. (Constant = Ka=3.50×10-8 )
HClO = OCl- + H+
_______ M
2.Estimate the initial [HClO] in an aqueous solution in which the [OCl-] is 0.0000167 M at equilibrium (make an approximate calculation assuming that initial concentration is equal to the equilibrium concentration).
(Constant = Ka=3.50×10-8 )
HClO = OCl- + H+
____________ M
3. Calculate the [OCl-] of a 2.27×10-3 M solution of the weak acid HClO (make an approximate calculation assuming that initial concentration is equal to the equilibrium concentration). Round your answer to 3 significant digits. (Constant = Ka=3.50×10-8 )
HClO = OCl- + H+
___________ M
1. Calculate the [HClO] in a solution if it has a pH 5.35. (Constant = Ka=3.50×10-8 )
HClO = OCl- + H+
_______ M
2.Estimate the initial [HClO] in an aqueous solution in which the [OCl-] is 0.0000167 M at equilibrium (make an approximate calculation assuming that initial concentration is equal to the equilibrium concentration).
(Constant = Ka=3.50×10-8 )
HClO = OCl- + H+
____________ M
3. Calculate the [OCl-] of a 2.27×10-3 M solution of the weak acid HClO (make an approximate calculation assuming that initial concentration is equal to the equilibrium concentration). Round your answer to 3 significant digits. (Constant = Ka=3.50×10-8 )
HClO = OCl- + H+
___________ M
Explanation / Answer
1. Calculate the [HClO] in a solution if it has a pH 5.35. (Constant = Ka=3.50×10-8 )
HClO = OCl- + H+
Solution :-
pH is given therefore using the pH lets calculate the concentration of the [H3O+]
pH= -log [H+]
[H+] = antilog [-pH]
= antilog [-5.35]
= 4.467*10^-7 M
[H+] = [OCl-] = 4.467*10^-6 M
Ka= [H+][ClO-] /[HClO]
Therefore
[HClO] = [H+][ClO-] / ka
= [4.467*10^-6] [4.467*10^-6] / 3.50*10^-8
= 5.70*10^-4 M
Therefore the [HClO] = 5.70*10^-4 M
2) Estimate the initial [HClO] in an aqueous solution in which the [OCl-] is 0.0000167 M at equilibrium (make an approximate calculation assuming that initial concentration is equal to the equilibrium concentration).
(Constant = Ka=3.50×10-8 )
Solution :- [H+] = [OCl-] = 1.67*10^-5 M
Ka= [H+][ClO-] /[HClO]
Therefore
[HClO] = [H+][ClO-] / ka
= [1.67*10^-5] [1.67*10^-5] / 3.50*10^-8
= 7.97*10^-3 M
Therefore the [HClO] = 7.97*10^-3 M
3) Calculate the [OCl-] of a 2.27×10-3 M solution of the weak acid HClO (make an approximate calculation assuming that initial concentration is equal to the equilibrium concentration). Round your answer to 3 significant digits. (Constant = Ka=3.50×10-8 )
Solution :-
Ka= [H+][ClO-] /[HClO]
3.50 *10^-8 = [x][x]/[2.27*10^-3]
3.50 *10^-8*[2.27*10^-3] = x^2
7.945*10^-11 =x^2
Taking square root of both sides we get
8.91*10^-6 M= x = [OCl-]
Therefore the concentration of the [OCl-]= 8.91*10^-6 M
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