Here below, there is a question and its answer. Where you can see many asterisks, it is my problem. What base does that kb belong to?? For goodness' sake, help!
"The Ka of HClO is 4*10^-8. What is the pH of a 0.02M solution of NaClO?"
When NaClO is dissolved in water, it produces 2 ions, Na+1 and ClO-1
ClO-1 is the negative ion from HClO. Hypochlorous acid is a weak acid, because the ClO-1 ion has a strong bond for the H+1. When the ClO-1 ion is dissolved in water, it attracts an H+1 from the water leaving an OH-1 ion. This produces a slightly basic solution.
NaClO → Na+1 + ClO-1
ClO-1 + H2O ↔ HClO + OH-1
The Na+1 ion is a spectator ion. It does not react with the water. So, it is not part of the equilibrium equation.
We use the pKa of HClO, because it is part of the equilibrium equation
http://en.wikipedia.org/wiki/Hypochlorou…
The numbers below are from the website above.
pKa = 7.53
Ka = 10^-7.53 = 2.95 * 10^-8
We need Kb, because of the OH-1 ion.
Kb = 1 * 10^-14 ÷ 2.95 * 10^-8 = 3.39 * 10^-7
Kb = [HClO] * [OH] / [ClO-] = 3.39 * 10^-7 ***************************************
Kb = x^2/ 0.02 = 3.39 * 10^-7
x = (0.02 * 3.39 * 10^-7)^0.5 = 8.23 * 10^-5
[OH-] = 8.23 * 10^-5
pOH = -1 * log 8.23 * 10^-5 = 4.08
pH = 14 – 4.08 = 9.92
"The Ka of HClO is 4*10^-8. What is the pH of a 0.02M solution of NaClO?"
When NaClO is dissolved in water, it produces 2 ions, Na+1 and ClO-1
ClO-1 is the negative ion from HClO. Hypochlorous acid is a weak acid, because the ClO-1 ion has a strong bond for the H+1. When the ClO-1 ion is dissolved in water, it attracts an H+1 from the water leaving an OH-1 ion. This produces a slightly basic solution.
NaClO → Na+1 + ClO-1
ClO-1 + H2O ↔ HClO + OH-1
The Na+1 ion is a spectator ion. It does not react with the water. So, it is not part of the equilibrium equation.
We use the pKa of HClO, because it is part of the equilibrium equation
http://en.wikipedia.org/wiki/Hypochlorou…
The numbers below are from the website above.
pKa = 7.53
Ka = 10^-7.53 = 2.95 * 10^-8
We need Kb, because of the OH-1 ion.
Kb = 1 * 10^-14 ÷ 2.95 * 10^-8 = 3.39 * 10^-7
Kb = [HClO] * [OH] / [ClO-] = 3.39 * 10^-7 ***************************************
Kb = x^2/ 0.02 = 3.39 * 10^-7
x = (0.02 * 3.39 * 10^-7)^0.5 = 8.23 * 10^-5
[OH-] = 8.23 * 10^-5
pOH = -1 * log 8.23 * 10^-5 = 4.08
pH = 14 – 4.08 = 9.92
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Ka = [H+][ClO-] / [HClO]
the Ka for HClO = 2.88x10^-8 so it would be better to use the Kb which is 3.47x10^-7
the original question gives the wrong Ka for HClO which is 2.88x10^-8
anyway, the Kb belongs to the base ClO- because when it goes into solution, ClO- takes the H+ from water leaving [OH-]. ClO- is the conjugate base to HClO
the Ka for HClO = 2.88x10^-8 so it would be better to use the Kb which is 3.47x10^-7
the original question gives the wrong Ka for HClO which is 2.88x10^-8
anyway, the Kb belongs to the base ClO- because when it goes into solution, ClO- takes the H+ from water leaving [OH-]. ClO- is the conjugate base to HClO
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Science is too hard