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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.

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https://brainmass.com/chemistry/acids-and-bases/concentration-without-hasselbach-542276

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 = ...

Solution Summary

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

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See Also This Related BrainMass Solution

Determine pH of a Solution: Henderson HasselBach Equations

Using a ICE table (initial change equilibrium)

a) Calculate the pH of solution that is 0.175M in acidic acid(HC2H3O2) ans 0.110Min potassium Acetate(KC2H302)

b) Perform same calculation using the Henderson HasselBach equations to determine pH of the solution

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