A 100.0 mL sample of a solution that is 0.125 M in HClO and 0.130 M in NaClO has added to it 2.00 mL of 0.100 M HCl. The Ka for HClO is 2.9 x 10-8. What is the pH of the resulting solution?

Start by finding the quantity of each species in the #"NaClO"# buffer immediately after the addition of #"HCl"#.

• #n("HCl") = c * V = 2.00 xx 10^(-4) color(white)(l) mol#
• #n("HClO") = c * V = 1.25 xx 10^(-2) color(white)(l) mol#
• # n("NaClO") = c * V = 1.30 xx 10^(-2) color(white)(l) mol#

Hydrochloric acid #"HCl"# would react with #"NaClO"# to produce a weak electrolyte #"HClO"# through the following double decomposition reaction:

#stackrel((color(grey)("strong")))("HCl"(aq)) + "NaClO" (aq) color(navy)(to) stackrel((color(grey)("weak")))("HClO"(aq)) + "NaCl"(aq)#

#"NaClO"# is in excess and therefore

#n("HClO","final") = n("HClO","initial")+ n("HCl") #
#color(white)(n("HClO","final")) = 1.2502 xx 10^(-2) color(white)(l) mol#

#n("NaClO", "final") = n("NaClO", "initial") - n("HCl")#
#color(white)(n("NaClO","final")) = 1.2998 xx 10^(-2) color(white)(l) mol#

The rest of #"NaClO"# completely disassociates to produce #"Na"^(+)# and #"ClO"^(-)#.

#"NaClO"(s) color(navy)(to) "Na"^(+)(aq) + "ClO"^(-)(aq)#

#n("ClO"^(-)) = n("NaClO") = 1.30 xx 10^(-2) color(white)(l) mol#

The hydrolysis of aqueous #"ClO"^(-)# from #"NaClO"# might pose an influence on the acidity (in other words, #"pH"#) of the solution. A "RICE" table can help with the derivation of the equilibrium proton concentration due to this reversible process.

R #" ClO"^(-)(aq) + "H"_2"O"(l) color(darkgreen)(rightleftharpoons) "OH"^(-) (aq) + "HClO"(aq)#
I #color(purple)(color(white)(ll)1.2998 xx 10^(-2) color(white)(-------) color(white)(ll)1.2502 xx 10^(-2)#
C #color(darkgreen)(" "-x color(white)(-------) +x color(white)(---) +x)#
E #color(navy)(color(white)(ll)1.2998 xx 10^(-2) color(white)(-------)1.2502 xx 10^(-2))#
#color(navy)(color(white)(--) -x color(white)(-----lll-) +x color(white)(---)+x)#
(all values in this table are in #mol#)

#K_a = 2.9 xx 10^(-8)# for #"HClO"# such that
#K_b = (1.0 xx 10^(-14))/ (2.9 xx 10^(-8)) = 3.448 xx 10^(-7)# for #"ClO"^(-)#, its conjugate base.

#(n("OH"^(-)) * n("HClO")) / (n("ClO"^(-))) = (["OH"^(-)] * ["HClO"]) / (["ClO"^(-)]) * V= K_b * V = 3.4 xx 10^(-7) * 0.102#

Where #n("OH"^(-))#, #n("HClO")#, and #n("ClO"^(-))# the #color(navy)("quantities")# (in number of moles) of each of the respective species present in the solution at its #color(navy)("equilibrium")# position. With reference to the RICE table,

#((x) * (1.2502 xx 10^(-2) + x)) / (1.2998 xx 10^(-2) - x) = 3.448 xx 10^(-7) * 0.102#

Solving this equation for #x# yields #x = 3.6color(grey)(56) xx 10^(-8) color(white)(l) color(purple)(mol)#. Therefore #n("OH"^(-)) = x = 3.6color(grey)(56) xx 10^(-8) color(white)(l) color(purple)(mol)# of #"OH"^(-)# would eventually be present in the solution as it comes to its equilibrium state.

The final solution has a volume of #100.0 + 2.00 = 102.0 color(white)(l) ml# (assuming no volume change due to the mixing.) Thus #["OH"^(-)] = n / V = 3.584 xx 10^(-7) color(white)(l) mol# at equilibrium.

#"pOH" = -log["OH"^(-)] = 6.445#
and therefore
#"pH" = "pKw" - "pOH" = 7.554#.