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#?

Question

Avery Adams · Accepted 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.

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#. Consequently, #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#. Thus, as a last 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#. Entering the buffer equation again: #pH=pK_a+log_10(((0.251*mol*L^-1))/((0.50*mol*L^-1)))# #=9.30+(-0.3)# #=9.0# AS REQUIRED....................

Henry Antrim · Answer

undefined To maintain a pH of 9.0 in a 20 mL volume of NH₃(aq) at 0.50 mol/L concentration, you would need to add 0.056 grams of ammonium chloride.

Ellie Andrews · Answer

undefined To calculate the amount of ammonium chloride (NH4Cl) needed to maintain a pH of 9.0 in a 20 mL volume of NH3(aq) at a concentration of 0.50 mol/L, you can use the Henderson-Hasselbalch equation. First, you need to determine the initial concentration of ammonia ($ \text{NH}_3 $) and ammonium ion ($ \text{NH}_4^+ $) using the given concentration of $ \text{NH}_3 $. Then, set up the equilibrium equation and solve for the concentration of $ \text{NH}_4^+ $ needed to achieve a pH of 9.0. Finally, calculate the amount of NH4Cl needed to achieve this concentration of $ \text{NH}_4^+ $.