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# Acid Base Equilibrium Questions

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1) A 0.223 molar solution of the weak acid, HA, with a pKa of 3.954 is titrated with a 0.142 molar solution of NaOH. What is the pH of the solution at the equivalence point of this titration?

2) 0.1405 mol of potassium cyanide (KCN) is used to prepare 609.1 mL of an aqueous solution. Take the pKa of hydrogen cyanide as 9.38. What's the pH of the solution?

3) A 0.297 molar solution of the weak acid, HA, with a pKa of 5.441 is titrated with a 0.327 molar solution of NaOH. What is the pH of the solution at the equivalence point of this titration?

4) The value of Kw for water at a temperature of 10oC is 8.5 x 10-15. The pH of completely pure (distilled) water at this temperature is:

5) What's the equilibrium constant, Kc, for the reaction...
2H2O+H30^+ + OH^- at 25 C?

6) What's the [H3O^+(aq)] in 1 x 10-10 M HCL at 25 C?

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1) A 0.223 molar solution of the weak acid, HA, with a pKa of 3.954 is titrated with a 0.142 molar solution of NaOH. What is the pH of the solution at the equivalence point of this titration?

There's something a little tricky here. There's no mention of volumes. What to do? Just put any volume of the weak acid we want, and go from there.

Therefore, let's just say that we have 1.000 L of HA that we are titrating with the 0.142 M NaOH.

What is the reaction?

NaOH + HA --------> NaA + H2O

In 1.000 L, how many moles of HA do we have at the start? Ans: 0.223 moles.

Therefore, how many moles of NaOH, do we need at equivalence? Ans: 0.223 moles.

What will be the volume of the 0.223 moles of NaOH that we add?

0.223 moles / x L = 0.142 M

Therefore, volume = 1.57 L of NaOH

Now what?

We need to find the total volume of the solution at equivalence so that we can calculate the concentation of the chemical species present.

Since we started with 1.000 L and added 1.57 L, the total volume at equivalence is 2.57 L

Of course, there won't be any HA at equivalence, for a reaction of a strong base with a weak acid goes to completion. But there will be NaA, the salt.

Therefore,

[NaA] = 0.223 moles / 2.57 L = 0.0868 M

Now what? We find out the pH of a 0.0868 M NaA solution.

This is where the pKa of 3.954 comes in.

A- + H2O <-----> HA + OH-

(We can just ignore the Na+)

Since there's OH- produced, the solution will be slightly basic.

Kb = [HA] [OH-] / [A-]

What's ...

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