The Ka of trichloroaceticacid(cl3CCOOH)is [cl3CC00^-][H+]/[cl3CC00H] = 0.10M in
ID: 929111 • Letter: T
Question
The Ka of trichloroaceticacid(cl3CCOOH)is [cl3CC00^-][H+]/[cl3CC00H] = 0.10M in water. Consider an aqueous solution containing neutral trichloroaceticacid and the cl3ccoo-anion in equilibrium What is the pKa of cl3cCOOH? What is the hydrogen ion concentration [H+], at pH 2? What is the ratio of the concentrations of trichloroaceticacid ion to trichloroaceticacid. [CL3CCOOO-]/[CL3CC00H]. atpH 2? THE Ka of trichloroaceticacid (CH_3CH_2)3NH+is (CH_3CH_2)3NH+ [H+]/[ (CH_3CH_2)3NH+] 10^-11M in water. Consider an aqueous solution containing neutral trifluoropropanoic acid and the (CH_3CH_2)3N an ion in equilibrium. What is the hydrogen ion concentration [H+], at Ph10? What is the ratio of die concentrations of trifluoropropanoate ion to (CH_3CH_2)3N/(CH_3CH_2)3NH+at pH 10?Explanation / Answer
Answer – IV) Given, Ka for trichloroacetic acid = 0.10 M
A)We know the pKa for Cl3CCOOH
We know formula
pKa = -log Ka
= - log 0.10
= 1.0
B) Given, pH = 2.0 , it is strong acid, since Ka value is big
We know,
pH = -log [H+]
so, [H+] = 10-pH
= 10-2.0
= 1.0*10-2 M
C) We know Henderson Hasselbalch equation
pH = pKa + log [Cl3CCOO-] / [Cl3CCOOH]
2.0 = 1.0 + log [Cl3CCOO-] / [Cl3CCOOH]
So, log [Cl3CCOO-] / [Cl3CCOOH] = 2.0-1.0
= 1.0
Taking anitlog from both site
[Cl3CCOO-] / [Cl3CCOOH] = 10
III) A) Given, Ka for triethylammonium ion = 1.0*10-11 M
At pH = 10, [H+] =?
So, pH = -log [H+]
[H+] = 10-pH
= 10-10.0
= 1.0*10-10 M
B) We know Henderson Hasselbalch equation
First we need to calculate the pKa
pKa = -log Ka
= -log 1.0*10-10
= 10
pH = pKa + log [(CH3CH2)3N] / [(CH3CH2)3NH+]
10.0 = 10.0 + log [(CH3CH2)3N] / [(CH3CH2)3NH+]
So, log [(CH3CH2)3N] / [(CH3CH2)3NH+]= 10.0-10.0
= 0.0
Taking antilog form both side
[(CH3CH2)3N] / [(CH3CH2)3NH+] = 1.0
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