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# Concentration without Hasselbach

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Please find the solutions without using the Hasselbach equation. Please show all calculations.

Question 1

Find [H^+] in a 25 degrees Celsius solution prepared by dissolving 0.20 moles of NH4Cl in water and diluting to a volume of 400 mL, given that Kb = 1.8 x 10^-5 for NH3 (aq) at 25 degrees Celsius.

Question 2

Find [H^+] in an aqueous 25 degrees Celsius that is .20 M in HC2H3O2 and .48 M in NaC2H3O2. For HC2H3O2, Ka = 1.8 x 10^-5 at 2 degrees Celsius.

Question #3

A solution is prepared by adding HCl and HC2H3O2 to water. In the equilibrium constant.

Ka = [H^+] [C2H3O2^-]/[HC2H3O2]

Which of the following is correct

a) The H^+ in the numerator includes only the H+ that comes from the ionization of the HC2H3O2.

b) The H+ in the numerator in includes both the H+ that comes from the ionization of the HC2H3O2 and from the ionization of the HCl.

##### Solution Summary

Three equilibrium problems that deal with finding out unknown concentrations are solved without using the Henderson-Hasselbach equation.

##### Solution Preview

Q1. NH4+ -----> NH3 + H+

[NH4+] = 0.20mol / 0.4L = 0.5M

If' x' is the change in concentration, then

Ka = 1 x 10^(-14) / Kb = [NH3][H+] / [NH4+] => 5.56 x 10^-10 = x^2/ (0.5 -x)
Hence 2.78 x 10^-10 - 5.56 x 10^-10x = ...

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