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tua28494
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Here is a problem I was working on but I do not have a solutions manual because Its an old textbook. I would just like someone to look over my work.
Calculate the pH of a 0.67 M NaClO solution.
NaClO -> Na+ + ClO-
ClO- + H2O -> HCLO + OH-
Kw = 1.0*10^-14
Ka HCLO = 3.5*10^-8
Could you please check if these constants are correct.
Kb = Kw/Ka HCLO
pH = -log [H+]
Kb = Kw/Ka HCLO = 1.0*10^-14/3.5*10^-8 = 2.86 * 10^-7
With Kb we can solve concentrations for OH-
Kb = [HClO] [OH-] / [ ClO-]
2.86 *10^-7 = (x)(x) / (0.67-x) Quadratic Formula
x = 4.38 * 10 -4 M
[OH-] = 4.38 * 10^-4 M
[H+] [ OH-] = 1.0 *10^-14
[H+] = 1.0 *10^-14 / 4.38 * 10^-4 M
= 2.28 * 10^-11 M
pH = - log[H+]
= - log( 2.28 * 10^-11)
= 10.64
This is a reasonalbe result.
My last question is how do I turn off overtype when I am preparing a new thread?
Thank you for your time.
Homework Statement
Calculate the pH of a 0.67 M NaClO solution.
NaClO -> Na+ + ClO-
ClO- + H2O -> HCLO + OH-
Kw = 1.0*10^-14
Ka HCLO = 3.5*10^-8
Could you please check if these constants are correct.
Homework Equations
Kb = Kw/Ka HCLO
pH = -log [H+]
The Attempt at a Solution
Kb = Kw/Ka HCLO = 1.0*10^-14/3.5*10^-8 = 2.86 * 10^-7
With Kb we can solve concentrations for OH-
Kb = [HClO] [OH-] / [ ClO-]
2.86 *10^-7 = (x)(x) / (0.67-x) Quadratic Formula
x = 4.38 * 10 -4 M
[OH-] = 4.38 * 10^-4 M
[H+] [ OH-] = 1.0 *10^-14
[H+] = 1.0 *10^-14 / 4.38 * 10^-4 M
= 2.28 * 10^-11 M
pH = - log[H+]
= - log( 2.28 * 10^-11)
= 10.64
This is a reasonalbe result.
My last question is how do I turn off overtype when I am preparing a new thread?
Thank you for your time.