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Buffer cacluation without Hasselbach

Find the solutions without using the Henderson-Hasselbalch equation. Please show all calculations.

Question #1

A student dissolves 0.25 moles of the acid HA in water and dilutes the solution to the final volume of 500 mL. The student then gives a small portion of the solution to her professor. The professor uses a pH meter and finds the solution given to him has a pH of 0.96

a) Find the H+ in the solution
b) Find the Ka (Ionization constant)

Question #2

A student mixes 3.0 mL of 0.20 M NH3 (aq) with 70.0 mL of 0.30 M NH4CL(aq). Find H+ in the solution Kb = 1.8 X10^-5.

Question #3

A student adds a certain number of moles of NH3 to 100 mL of a 0.4 MNH4CL(aq) solution and finds the resulting solution has H+ = 8.3 x 10^- M. How many moles of NH3 were added (assume the final solution volume is still 100 mL. Kb = 1.8 x 10^-5)?

Question #4

Which one of the following gives a buffer solution?

a) 50 Ml of 0.1 M HCl (aq) mixed with 80 mL of 0.1 M NaOH (aq).
b) 80 Ml of 0.1 M HCl (aq) mixed with 50 mL of 0.1 M NaOH (aq).
c) 50 Ml of 0.1 M HC2H3O2 (aq) mixed with 50 mL of 0.1 M NaOH (aq).
d) 50 Ml of 0.1 M HC2H3O2 (aq) mixed with 80 mL of 0.1 M NaOH (aq).
e) 80 Ml of 0.1 M HC2H3O2 (aq) mixed with 500 mL of 0.1 M NaOH (aq).
f) None of the above

Solution Preview

Q1. pH = -log[H+]. Hence [H+] = 10^(-pH) = 10^(-0.96) = 0.11M

Initial concentration of HA = 0.25 mol / 0.5L = 0.5M

Ka = [H+][A-] / [HA] = 0.11 x 0.11 / (0.5 - 0.11) = 3.10 x 10^(-2)

Q2. [NH3] = (0.003 x 0.2) / 0.073 = 0.00822M

[NH4+] = 0.07 x 0.30 / 0.073 = 0.288M

NH3 + ...

Solution Summary

A few buffer problems are solved without using the Henderson-Hasselbach equation.

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