a. What is the pH of a 1.08 M solution of (CH 3 CH 2 ) 2 NH that is also 0.49 M
ID: 712487 • Letter: A
Question
a. What is the pH of a 1.08 M solution of (CH3CH2)2NH that is also 0.49 M in diethylammonium sulfate, the salt of it's conjugate acid? The Kb for (CH3CH2)2NH = 6.9 X 10-4.
b. In one flask you are given 160.0 mL of a 6.20 M solution of hypochlorous acid. In a separate flask you are given 280.0 mL of a 0.70 M solution of gallium hypochlorite. What is the final pH when these two flasks are mixed together and allowed to equilibrate? The Ka for hypochlorous acid = 2.9 X 10-8
c. What is the final pH when 3.20 mL of 1.05 M perchloric acid is added to 110.00 ml the buffer in problem 1 above? The Ka for CH3CH2CH2COOH is 1.3 X 10-5. Hints: look at you buffer, which is the acid, which is the base? What type of reagent did you just add?
Explanation / Answer
a) given: mixture of 1)1.08 M solution of (CH3CH2)2NH + 2)0.49 M in [(CH3CH2)2NH2]2SO4
[(CH3CH2)2NH2]2SO4 ---->2(CH3CH2)2NH2+ + SO42-
concentration of (CH3CH2)2NH2+=[(CH3CH2)2NH2+]=2*0.49=0.98
pH=pka+log [base]/[acid] (Henderson-hasselbach equation)
Kb (CH3CH2)2NH = 6.9 *10-4.
ka=kw/kb=(10^-14)/(6.9*10^-4)=1.449*10^-11
pka=-logka=-log (1.449*10^-11)=10.8
pH=10.8+log[(CH3CH2)2NH]/[(CH3CH2)2NH2+]=10.8+log(1.08/0.98)=10.8
pH=10.8
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