Consider the titration of #50.0 mL# of #0.20 M# #NH_3# Kb #1.8x10^-5# with #0.20 M# #HNO_3#. How do you calculate the pH after addition of 50.0mL of the titrant?

Question

Luna Chen · Accepted Answer

You can do it like this:

The equation for the neutralisation is: #sf(NH_3+H^+rarrNH_4^+)# We can find the number of moles using #sf(n=cxxv)#: #sf(n_(NH_3)=0.20xx50.0/1000=0.01)# #sf(n_(H^+)=0.20xx50.0/1000=0.01)# This tells us that: #sf(n_(NH_4^+)=0.01)# #sf(c=n/v)# The total volume is 50.0 + 50.0 = 100.0 ml #:.##sf([NH_4^+]=(0.01)/(100.0/1000)=0.10color(white)(x)"mol/l")# The ammonium ion partially dissociates: #sf(NH_4^+rightleftharpoonsNH_3+H^+)# For which: #sf(K_a=([NH_3][H^+])/([NH_4^+])# To find the #sf(pH)# we need to know the value of #sf(K_a)#. We can get this by using the expression: #sf(K_axxK_b=K_w=10^(-14)color(white)(x)"mol"^2."l"^-2)# at #sf(25^@C)# #:.##sf(K_a=10^(-14)/(1.8xx10^(-5))=5.55xx10^(-10)color(white)(x)"mol/l")# Now we set up an ICE table based on concentrations: #sf(" "NH_4^+" "rightleftharpoons" "NH_3" "+" "H^+)# #sf(color(red)(I)" "0.10" "0" "0")# #sf(color(red)(C)" "-x" "+x" "+x)# #sf(color(red)(E)" "(0.1-x)" "x" "x)# #:.##sf(K_a=x^2/(0.1-x))# Because the value of #sf(K_a)# is so small we can assume that the size of #sf(x)# is negligible compared with #sf(0.1)# so we can write: #sf(K_a=x^2/0.1=5.55xx10^(-10))# #:.##sf(x^2=5.55xx10^(-10)xx0.1=5.55xx10^(-11))# #:.##sf(x=[H^+]=sqrt(5.55xx10^(-11))=7.449xx10^(-6)color(white)(x)"mol/l")# #sf(pH=-log[H^+]=-log[7.449xx10^(-6)]=5.12)# #sf(color(red)(pH=5.12))# This shows that the pH is slightly acidic at the equivalence point which is typical of a salt produced from a strong acid and a weak base.

Kai Bowman · Answer

undefined To calculate the pH after the addition of 50.0 mL of the titrant (0.20 M HNO3) to 50.0 mL of 0.20 M NH3 (aqueous ammonia), you would first determine the number of moles of NH3 and HNO3 reacted.

Since NH3 is a weak base and HNO3 is a strong acid, the reaction between them will go to completion. The moles of NH3 initially present is given by:

moles of NH3 = (volume of NH3 solution in L) * (molarity of NH3)

Then, you'll need to determine the limiting reactant and calculate the moles of excess reactant remaining.

After that, you'll calculate the concentrations of NH3 and NH4+ (the conjugate acid of NH3) in the final solution. Use the stoichiometry of the reaction to determine how many moles of NH3 were converted to NH4+.

Finally, you'll use the equilibrium expression for the reaction of NH3 with water to find the concentration of OH- ions, and then use that to calculate the pOH. From the pOH, you can find the pH of the solution using the relationship: pH = 14 - pOH.