How much ammonium chloride do we need to add to a #20*mL# volume of #NH_3(aq)# at #0.50*mol*L^-1# concentration, to maintain a #pH=9.0#?

1 Answer
Mar 8, 2017

Answer:

We need (i) approx. #270*mg#, and (ii) we need to use the buffer equation: #pH=pK_a+log_10(([NH_4^+])/([NH_3]))#.

I do not think the given answer is kosher.

Explanation:

Now we know that #pK_a+pK_b=14# for an aqueous solution under standard conditions, #K_b=2xx10^-5#, and thus #pK_b=4.70#, and #pK_a=9.30#

And we substitute the #pK_a# value back into the original equation, with the desired #pH#.

And thus,

#9=9.30+log_10(([NH_4^+])/([NH_3]))#

And thus #-0.30=log_10(([NH_4^+])/([NH_3]))#

#10^(-0.3)=([NH_4^+])/([NH_3])#

And we know that #[NH_3]=0.50*mol*L^-1#.

So #[NH_4^+]=10^(-0.3)xx0.50*mol*L^-1=0.25*mol*L^-1#

Now #"concentration"# #=# #("Moles of solute")/("Volume of solution")#, but the volume of solution was SPECIFIED to be #20xx10^-3L#.

So #"moles of solute"=20xx10^-3cancelLxx0.25*mol*cancel(L^-1)#

#=5.01xx10^-3*mol# of #NH_4Cl#; and this represents a mass of

#5.01xx10^-3*molxx53.49*g*mol^-1=0.268*g#.

So a final check,

#[NH_4^+]=((0.268*g)/(53.49*g*mol^-1))/(20xx10^-3*L)=0.251*mol*L^-1#.

#[NH_3]=0.50*mol*L^-1#.

Resubstituting into the buffer equation:

#pH=pK_a+log_10(((0.251*mol*L^-1))/((0.50*mol*L^-1)))#

#=9.30+(-0.3)#

#=9.0# AS REQUIRED....................