SilverBandCry! said:
I am confused about this question -->
You have 1 Liter of 0.25 propanoic acid (CH3CH2COOH).
1. Calculate the pH of this solution.
2. Add 0.10 moles of NaOH to this solution. Calculate the resulting pH.
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--When I solved this I got 5.5 for number 1, and 4.75 for number 2 but I think it is wrong because if you add base, shouldn't the pH increase?
thanks
First you need either the pKa or Ka of CH3CH2COOH.
Since CH3CH2COOH is a weak acid it does not completely dissociate. So at equil. there will be less than 0.25 moles of H+. To solve #1
CH3CH2COOH <---> CH3CH2COO- + H3O+
Ka = [H3O+][CH3CH2COO-] / [CH3CH2COOH]
Lets assume only x moles of CH3CH2COOH have dissociated at equil. Then the equil. conc of CH3CH2COOH will be (0.25 - x)moles. From the balanced equation above the mole ratio of CH3CH2COOH to both CH3CH2COO- and H3O+ is 1 to 1. So if x moles of CH3CH2COOH have dissiocated then x moles of CH3CH2COO- and x moles H3O+ have been formed.
So
Ka = (x.x)/ (0.25 - x) which is approximately (x.x)/0.25
If we assume x << 0.25 (i.e x is so much smaller than 0.25 that it wont really matter in the denominator)
So with the Ka value you can solve for x.
For #2
Realise that adding the NaOH (base) will effectively neutralize the CH3CH2COOH. Also by Le Chatlier's principle, it will favor the forward reaction and thereby cause the complete dissiocation of CH3CH2COOH if enough NaOH is added. Since you have only 0.1 moles of NaOH the solution should still be slightly acidic.
Since NaOH is a strong base it will completely dissociate and produce 0.10 moles of OH-. So 0.1mole of your CH3CH2COOH will be neutralized. Thus you will have (0.25 - 0.10 = 0.15) moles of unneutralized CH3CH2COOH. This 0.15 will partially dissociate and with the Ka you repeat the calculations from #1 to get the new pH of the remaining solution. So your [H+] now should be smaller than 0.15 moles.
