Working with Acid base equlibria
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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.© BrainMass Inc. brainmass.com March 6, 2023, 12:49 pm ad1c9bdddf
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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)
All questions answered with proper expanations. Answer as a Word attachment.