Calculate the pH of a solution of 0.10 M #HF# that is also 0.15 M #HCN# if #K_a# for #HF# is #1.0xx10^-5# and #K_a# for #HCN# is #1.0xx10^-7#?

Question

Luna Chen · Accepted Answer

#"pH" ~~ 2.09# In principle, it does not matter much which acid you allow to go first, as long as the smaller #K_a# is smaller by at least #10^3#. You may want to try this in reverse order, doing #"HCN"# first and #"HF"# second, just to see if you get about the same #"pH"# again. (You should get #x = 9.64 xx 10^(-6) "M"# for #["H"_3"O"^(+)]# from #"HCN"# going first, and #x ~~ "0.00812 M"# for #"HF"# going second. However, when #"HF"# goes, the small #x# approximation can only be done with #"0.10 M"#, and you can say that #9.64 xx 10^(-6)# #"<<"# #x# instead in this case.) Also, a common pitfall is to forget about the #"H"_3"O"^(+)# contributed by the acid that goes first. That should be included in the initial concentration of #"H"_3"O"^(+)# for the acid that goes second, because they suppress each other. Since these are both weak acids that "have" similar #K_a# values, the weaker one (#"HCN"#) is not supposed to be ignored. But these #K_a# values are not correct... #K_a("HF") = 6.6 xx 10^(-4)# #K_a("HCN") = 6.2 xx 10^(-10)# So, we'd "have" to do sequential ICE tables, but we should not have to for this problem. We SHOULD be able to just ignore #"HCN"#... For #"HF"#: #"HF"(aq) + "H"_2"O"(l) rightleftharpoons "F"^(-)(aq) + "H"_3"O"^(+)(aq)# #"I"" "0.10" "" "" "-" "" "0" "" "" "" "0# #"C"" "-x" "" "" "-" "+x" "" "" "+x# #"E"" "0.10-x" "-" "" "x" "" "" "" "x# Therefore, using its #K_a# (which is actually #6.6 xx 10^(-4)#): #6.6 xx 10^(-4) = (["F"^(-)]["H"_3"O"^(+)])/(["HF"])# #= x^2/(0.10 - x)# We do the small #x# approximation because that was the intent of the problem, and: #6.6 xx 10^(-4) ~~ x^2/0.10# #=> x = sqrt(0.10 cdot6.6 xx 10^(-4)) = "0.00812 M" = ["H"_3"O"^(+)]_("HF")# (the true answer was #"0.00780 M"#, #4.15%# error, which is just within range of good enough.) We will discover that concentration does not vary significantly as it becomes the initial in the subsequent ICE table. For #"HCN"# after #"HF"#: #"HCN"(aq) + "H"_2"O"(l) rightleftharpoons "CN"^(-)(aq) + "H"_3"O"^(+)(aq)# #"I"" "0.15" "" "" "-" "" "" "0" "" "" "" "0.00812# #"C"" "-x" "" "" "-" "" "+x" "" "" "+x# #"E"" "0.15-x" "-" "" "" "x" "" "" "" "0.00812+x# For this #K_a# (which is actually #6.2 xx 10^(-10)#): #6.2 xx 10^(-10) = (["CN"^(-)]["H"_3"O"^(+)])/(["HCN"])# #= (x(0.00812+x))/(0.15 - x)# Here we see that #6.2 xx 10^(-10)# #"<<"# #0.00812#, or #8.12 xx 10^(-3)#, as well as #0.15#. Therefore, in the presence of #"HF"#: #6.2 xx 10^(-10) = (x(0.00812))/(0.15)# #= 0.0541x# #=> x = 1.15 xx 10^(-8) "M" = ["H"_3"O"^(+)]_("HCN")# As a result, the total #["H"_3"O"^(+)]# would be: #color(blue)(["H"_3"O"^(+)]) = ["H"_3"O"^(+)]_("HF") + ["H"_3"O"^(+)]_("HCN")# #= "0.00812 M" + 1.15 xx 10^(-8) "M" ~~ color(blue)("0.00812 M")# (Clearly, nothing much changed, because #"HCN"# is a much weaker acid in reality, by a factor of about #1000000#.) That makes the #bb"pH"#: #color(blue)("pH" = -log(0.00812) = 2.09)#

Mateo Aguilar · Answer

undefined To calculate the pH of the solution containing both HF and HCN, we need to consider the dissociation of both acids and the subsequent equilibrium concentrations of their respective conjugate bases.

1. Start by writing the dissociation equations for HF and HCN:

HF (aq) ⇌ H^+ (aq) + F^- (aq)
HCN (aq) ⇌ H^+ (aq) + CN^- (aq)

2. Use the given Ka values to set up equilibrium expressions:

Ka for HF = [H^+][F^-]/[HF] = 1.0 × 10^-5
Ka for HCN = [H^+][CN^-]/[HCN] = 1.0 × 10^-7

3. Since HCN is a weaker acid (larger Ka value) than HF, we can assume that the dissociation of HCN is negligible compared to HF. Therefore, we can approximate the concentration of H^+ ions mainly coming from the dissociation of HF.

4. Calculate the equilibrium concentration of H^+ ions using the Ka value for HF and the initial concentration of HF:

[H^+] = sqrt(Ka * [HF])

5. Use the concentration of H^+ ions to calculate the pH of the solution:

pH = -log[H^+]

Plug in the calculated value of [H^+] into the pH equation to find the pH of the solution.