Question

1.The pH of a 2.65×10^-3 M solution of a weak base is 9.17. Calculate pK_b 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.

Answer 1

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)    $\rightleftharpoons$ 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  

K_b is given as

K_b = [ 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

K_b = (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

pK_b = - log K_b

= - log 8.26 x 10 ^-08

pK_b   = 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