Question

Calculate the percent ionization of 0.135 M lactic acid (Ka=1.4×10−4). Calculate the percent ionization of 0.135 M lactic acid in a solution containing 8.0×10⁻³  M sodium lactate.

How many grams of dry NH4Cl need to be added to 2.50 L of a 0.800 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.62? Kb for ammonia is 1.8×10−5.

Answer 1

1) Lactic acid+H2O↔lactate ion +H3O+

ICE table

	[lactic acid]	[H3O+]	[lactate]
initial	0.135M	0	0
change	-x	+x	+x
equilibrium	0.135-x	x	x

Ka=[lactate][H3O+]/[lactic acid]

Or,1.4*10^-4=x*x/(0.135-x)

Or, 1.4*10^-4=x*x/(0.135) [ignore x compared to 0.135 as dissociation of weak acid is very low and x is very small comparatively)

Or,X^2=0.189*10^-4

X=0.435*10^-2 M=0.0043M

So [lactic acid] ionized at equilibrium=x=0.0043M

%ionization=0.0043/0.135*100=3.185%

2)In sodium lactate solution or buffer

[lactate]=8.0*10^-3M=0.008M

ICE table

	[lactic acid]	[H3O+]	[lactate]
initial	0.135M	0	0.008M
change	-x	+x	0.008+x
equilibrium	0.135-x	x	0.008+x

Ka=[lactate][H3O+]/[lactic acid]

,1.4*10^-4=x*(x+0.008)/(0.135-x)

Or,1.4*10^-4=x*0.008/0.135 [x<<<0.008,0.135]

Or,x=0.0829*10^-4M

% ionization of lactic acid=0.0829*10^-4/0.135*100=0.00614%

3) How many grams of dry NH4Cl need to be added to 2.50 L of a 0.800 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.62? Kb for ammonia is 1.8×10−5.

Moles of NH3=2.50L *0.800 mol/L=2 moles

For a buffer solution, pOH=pkb+log [conjugate acid]/conjugate base] Henderson-hasselbach equation]

Kb=1.8*10^-5

Pkb=-log kb=4.74

pOH=14-8.62=5.38

5.38=4.74+log [acid]/2

Or,5.38-4.74=log[acid]/2

Or,0.64= log[acid]/2

Or,10^0.64=[acid]/2

Or4.365=[acid]/2

Or,[acid]=2*4.365=8.73 moles

Acid here is NH4+ ion or NH4Cl

Molar mass of NH4Cl=53.491g/mol

Mass of acid to be added=8.73 moles*53.491g/mol=466.97 g