1.The pH of a 2.65×10-3 M solution of a weak base is 9.17. Calculate pKb for this base to two decimal places.
2.Determine the mass (in g) of sodium butanoate
(NaC3H7COO) that must be added to 78.9 mL of
0.609 M butanoic acid to yield a pH of 6.43. Report your answer to
3 significant figures.
Assume the volume of the solution does not change and that the 5%
approximation is valid.
pOH of the solution is
pOH = 14 - pH
= 14-9.17
= 4.83
Now, pOH = - log [ OH- ]
[ OH- ]= 10 - pOH
= 10 - 4.83
= 1.48 x 10 -5 M
Weak base dissociate as ,BOH (aq) B +(aq) + OH -(aq)
initial concentration 2.65 x 10 -3 0 0
Change -X +X +X
equilibrium conc.n 2.65 x 10 -3 - X +X +X
Kb is given as
Kb = [ B +][ OH -] / [ BOH] = X x X / 2.65 x 10 -3 - X
X is very small as compare to 2.65 x 10 -3 so we can write 2.65 x 10 -3 - X as 2.65 x 10 -3
Kb = (1.48 x 10 -5) 2 / 2.65 x 10 -3
= 2.19 x 10 -10 / 2.65 x 10 -3
= 8.26 x 10 -08
pKb = - log Kb
= - log 8.26 x 10 -08
pKb = 7.08
2) mixture of butanoic acid and sodium butanoate is buffer solution . It's pH is calculated by using Henderson's equation
pH = pKa + log [ sodium butanoate ] / [ butanoic acid]
6.43 = 4.82 + log [ sodium butanoate ] / [ butanoic acid]
log [ sodium butanoate ] / [ butanoic acid] = 6.43- 4.82
= 1.61
log [ sodium butanoate ] -log [ butanoic acid] = 1.61
no of moles of butanoic acid= molarity x volume in litre = 0.609 x 0.0789=0.048 moles
log [ sodium butanoate ] -log 0.048 = 1.61
log [ sodium butanoate ] -( - 1.32 ) = 1.61
log [ sodium butanoate ] = 1.61- 1.32
= 0.29
[ sodium butanoate ] = 10 0.29
= 1.95 moles
Therefore mass of sodium butanoate = no of moles x molar mass
= 1.95 x 114 gm
= 223.44 gm
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