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Working with Acid base equlibria

Please show all work and remember this is a first year course. Chem is my weakest subject.
Please send responsse in Word.
Also, if you are not willing to show me the proper way to do these by showing the calculations and will only supply me with some notes tht I could read out of any textbook myself, please do not take this posting and my money as others have.

1. Determining Ka. What is the Ka value for a weak acid if a 0.0150-M solution has a pH of 2.67?

2. Calculations With Ka. What is the pH of a 0.250-M benzoic acid solution? Ka = 6.3 * 10^-5

3. Polyprotic Weak Acid. What are pH, [H3O+], [HSO3-], and [SO32-] in a 1.50 M sulfurous acid solution?

4. Calculations With Kb. What is the Kb value for a weak base if a 0.250-M solution has a pOH of 4.7?

5. The pH of a Salt Solution. What is the pH of 0.50-M sodium benzoate? The benzoate ion (C7H5O2-) is the anion of benzoic acid.

Solution Preview

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HA(aq) <---> H+(aq) + A-(aq)
The equilibrium constant for the dissociation of an acid is known as Ka. The larger the value of Ka, the stronger the acid.

That is, Ka = [hydrogen ions][acid ions]/[acid] -------(1)

So we have to write down the dissociation equation.

If one mole of the acid, on ionization, gives one mole each hydrogen and "acid ions" then,

Ka = [hydrogen ions] and the concentration of hydrogen ions is related to the pH as

pH = -log10 [H+] ------(2)

Determining Ka. What is the Ka value for a weak acid if a 0.0150-M solution has a pH of 2.67?

For weak acid also, Ka = [H+] [A-]/[HA]
Since pH is given, [H+] = 10-pH = 10-2.67 = 0.00214

We have the following reaction HA(aq) <---> H+(aq) + A-(aq)

Initial ...

Solution Summary

All questions answered with proper expanations. Answer as a Word attachment.

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