Pyridine , C 5 H 5 N , is a weak base that dissociates in water as shown above.
ID: 827089 • Letter: P
Question
Pyridine , C5H5N , is a weak base that dissociates in water as shown above. At 25 C , the base dissociation constant , Kb , for C5H5N is 1.7 *10-9 .
a) Determine the hydroxide ion concentration and the percentage dissociation of a 0.15 molar solution of pyridine at 25 C .
b) Determine the pH of a solution prepared by adding 0.0500 moles of solid C5H5NHCl to 100. millimeters of a 0.15-molar solution of C5H5N.
c) If a 0.0800 mole of solid magnesium chloride , MgCl2 is dissolved in the solution prepared in part (b) and resulting solution is well-stirred , will a precipitate of Mg(OH)2 form ? Show calculation to support your answer . ( Assume the volume of the Mg(OH)2 is 1.5 * 10-11 ).
Thank You!!!
Explanation / Answer
a.
C5H5N + H2O <=> HC5H5N+ + OH-
0.15 / 0 0
-x / +x +x
0.15-x / x x
In 0.15-x, x is insignificant
Kb = [HC5H5N+][OH-]/[C5H5N] = x^2/0.15 = 1.7 x 10^-9
x = 1.6 x 10^-5 M
[OH-] = 1.6 x 10^-5
% ionization = [HC5H5N+]eq / [C5H5N]initial
= 1.6 x 10^-5 / 0.15 x 100%
= 0.01 %
b.
If 0.05 mole is added to 100 mL
concentration = 0.05 mol / 0.1 L = 0.5 M
Hendersal Hasselbalch equation
pH = pKa + log ([base]/[acid])
Ka = Kw/Kb = 5.88 x 10^-6
pKa = -log Ka = 5.230448
acid = C5H5NHCl
base = C5H5N
pH = pKa + log ([base]/[acid])
= 5.23 + log (0.15 / 0.50)
= 4.71
c.
concentration of MgCl2 = [Mg2+] = 0.8 M
[OH-] = 10^-pOH = 10^-(14-pH)
= 10^-9.29
= 5.1 x 10^-10 M
Ksp = [Mg2+][OH-]^2 = 1.5 x 10^-11
TIP = (0.8)(5.1 x 10^-10)^2
= 2.08 x 10^-19
TIP < Ksp, thus there is no precipitation
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.