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ID: 636467 • Letter: T
Question
the sample space listing the eight the sample space listing the eight the sample space listing the eight 1. The overall dissociation of malonic acid, H:CaH:04, is represented below. The overall dissociation constant is also indicated. HCsH:04 2H+CHaO To a 0.025-molar solution of malonic acid, a strong acid is added until the pH is 1.5. Calculate the [CaH:01 in the resu solution. (Assume the change in volume is negligible.) The value of the second dissociation constant,K is 2.0x 10 K=3.0 x 10. . 5-0 X10.9 (3.0X10 (0.0316) X (CH3H204- . o.o25 03 204 0.036 ? CH+ ANSWER O IS M 2. Determine the pH of a 2.25 M solution of pyridinium nitrate (CsHsNHNOs) at 25 C [Pyridinium nitrate dissociates in water to give pyridinium ions (CsHsNH'), the conjugate acid of pyridine (Ko 1.73 x 109), and nitrate ions (NO3)] ANSWER:Explanation / Answer
Ans 2
The dissociation reaction
C5H5N = C5H5NH+ + OH-
Base dissociation constant
Kb = 1.73 * 10^-9
Acid dissociation constant Ka = Kw/Kb
= 10^-14/(1.73 * 10^-9 )
Ka = 5.78 * 10^-6
The dissociation reaction with ICE TABLE
C5H5NH+ = C5H5N + H+
I 2.25
C - x +x +x
E 2.25-x x. x
Equilibrium constant expression of the reaction
Ka = [C5H5N] [H+] / [C5H5NH+]
5.78 * 10^-6 = [x] [x] / [2.25 - x]
2.25 >> x
5.78 * 10^-6 = [x] [x] / [2.25 ]
x2 = 1.3*10^-5
x = 3.61 * 10^-3
[H+] = 3.61 * 10^-3 M
pH = - log [H+] = - log (3.61 * 10^-3)
= 2.44
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