Calculate the #"pH"# change in a buffer solution upon the addition of strong acids / base of known concentration?

Calculate the #"pH"# change that take place when a #100color(white)(l)"ml"# portion of A), B)
is added to a #40color(white)(l)"ml"# of buffer solution made up of

  • #0.2color(white)(l)"M"# #"NH"_3# and
  • #0.3color(white)(l)"M"# #"NH"_4"Cl"#

given that #"Kb"=1.8 × 10^(−5) color(white)(l) "mol"\cdot "dm"^(-3)# for #"NH"_3#. (Chemistry Libretext )

A) #0.5color(white)(l)"M"# #"NaOH"#
B) #0.05color(white)(l)"M"# #"HCl"#

1 Answer
Apr 7, 2018

Answer:

A) #Delta"pH"=+3.50#
B) #Delta"pH"=-0.76#

Explanation:

Initial #sf("pH")#
Start by constructing a #"RICE"# table (all figures are in #"mol"\cdot "dm"^(-3)#)
for the dissociation of #"NH"_3# (or #"NH"_3\cdot "H"_2"O"(aq)# to be precise)
Let the decrease in the concentration of ammonia, #"NH"_3\cdot "H"_2"O"(aq)#, be #x color(white)(l)"mol"\cdot "dm"^(-3)#.

Note that as a soluble ionic compound, #"NH"_4"Cl"# ionizes completely to produce #"NH"_4^(+)(aq)# and #"Cl"^(-)(aq)# when dissolved in water.
Hence the concentration of ammonium #["NH"_4^(+)(aq)]=["NH"_4"Cl"]=0.3color(white)(l)"mol"\cdot"dm"^(-3)# . The self-ionization of water is deemed neglectable.

#color(grey)R" " "NH"_3\cdot"H"_2"O" (aq)\rightleftharpoons "NH"_4^(+)(aq) + "OH"^(-)(aq)#
#color(grey)(I)color(white)(-ll-)0.008color(white)(----ll)0.012#
#color(grey)(C)color(white)(-ll-)-x color(white)(--ll--)+xcolor(white)(-l-)+x#
#color(grey)(E)color(white)(--)0.008-x color(white)(-ll-)0.012+x color(white)(-lll)+x#

Here (in order to simplify the calculation) it is assumed that at equilibrium #0.2-x ~~ 0.2# and #0.3-x ~~ 0.3# given that the value of #"K"_b# far smaller than that of either #["NH"_4^+]# or #["NH"_3\cdot "H"_2"O"]#.

#"K"_b=(["NH"_4^+]\cdot["OH"^-])/(["NH"_3\cdot "H"_2"O"])=(0.012\cdot x)/(0.008)#

Solving for #x# gives
#x=1.2*10^(-5)color(white)(l)"mol"\cdot"dm"^(-3)#
Hence #["OH"^-]=1.2*10^(-5)color(white)(l)"mol"\cdot"dm"^(-3)#,

#["H"^+]="K"_w/(["OH"^-])=8.3\cdot 10^(-10)color(white)(l)"mol"\cdot"dm"^(-3)#
assuming that #"T"=298 color(white)(l)K# and #"K"_w=1.0×10^(-14)#

Therefore
#ul("pH"_("initial")=-log["H"^+]=9.08)#

#"pH"# on the addition of #sf(100 color(white)(l)"dm"^(-3))# of #sf(0.5color(white)(l)"mol"\cdot"dm"^(-3))# #sf("NaOH")#
Addition of sodium hydroxide changes the concentration of #["OH"^-]# in the initial solution; one might expect the solution to stay basic as it reaches equilibrium. All #NH"_4^+# would have been converted to #"NH"_3# with some #"OH"^-# in excess. Thus it is possible to solve for the final #["OH"^-]# and hence #"pH"# using the same #"RICE"# table as in that the calculation of the initial #"pH"#.

#["OH"^(-)(aq)]=["NaOH"]-["NH"_4^+(aq)]=0.038color(white)(l)"mol"#

#color(grey)R" " "NH"_3\cdot"H"_2"O" (aq)\rightleftharpoons "NH"_4^(+)(aq) + "OH"^(-)(aq)#
#color(grey)(I)color(white)(-ll-)0.020color(white)(------)color(white)(-ll-)0.038#
#color(grey)(C)color(white)(-ll-)-x color(white)(--ll--)+xcolor(white)(-l-)+x#
#color(grey)(E)color(white)(-ll)0.020-x color(white)(----)x color(white)(--)0.038+x#

Similarly,
#(x\cdot 0.038)/(0.020)="K"_b#
#x=6.32\cdot 10^(-6)#

#"pH"=14+log["OH"^-]=12.58#

#ul(Delta"pH"=3.50)#

#"pH"# on the addition of #sf(100 color(white)(l)"dm"^(-3))# of #sf(0.05color(white)(l)"mol"\cdot"dm"^(-3))# #sf("HCl")#
All #["H"^+]# were consumed in this case, converting part of #"NH"_3# to #"NH"_4^+#. Similar to calculation of the initial #"pH"#,

#color(grey)R" " "NH"_3\cdot"H"_2"O" (aq)\rightleftharpoons "NH"_4^(+)(aq) + "OH"^(-)(aq)#
#color(grey)(I)color(white)(-ll-)0.003color(white)(----ll)0.017#
#color(grey)(C)color(white)(-ll-)-x color(white)(--ll--)+xcolor(white)(-l-)+x#
#color(grey)(E)color(white)(--)0.003-x color(white)(-ll-)0.017+x color(white)(-lll)+x#

#"K"_b=(["NH"_4^+]\cdot["OH"^-])/(["NH"_3\cdot "H"_2"O"])=(0.017\cdot x)/(0.003)#

#x=2.1*10^(-6)#

#"pH"=14+log["OH"^-]=8.32#

#ul(Delta"pH"=-0.76)#