Question #4c428

2 Answers
Apr 26, 2017

Answer:

When HCl is added into a #BaSO_4#(aq) solution, the conjugate base #(SO_4^-2)# adds a proton to => #(HSO_4^-2)# a weak acid and #BaSO_4# solubility is increased.

Explanation:

#H_2SO_4# is one of the 6 strong acids #(HCl, HBr, HI, HNO_3, HClO_4, H_2SO_4#). However, for #H_2SO_4#; 100% ionization occurs only in the 1st ionization step as the bisulfate is a weak acid #(HSO_4^(-))# with #K_a# = #1.1X10^-2#. This means that if an acid is added into a solution of sulfuric acid, the acidic hydrogen would bond with the sulfate ion => bisulfate ion. This essentially removes the #(SO_4^-2)# from the #BaSO_4# equilibrium causing more #BaSO_4# to dissolve and ionize in order to replace the #(SO_4^-2)# removed from the solution system.

Same Explanation but with equilibrium rxn of #BaSO_4#
Adding HCl => #H^+# into #BaSO_4#(s) #rightleftharpoons# "#Ba^(+2)# + #SO_4^(2-)#
=>#BaSO_4#(s) #rightleftharpoons# "#Ba^(+2)# + #HSO_4^(2-)# where #HSO_4^(2-)# is no longer a part of the #BaSO_4# equilibrium. This causes reaction to shift right to deliver more #SO_4^(2-)# to replace the sulfate converted to bisulfate. This process will continue until all of the added HCl is removed, or all of the Barium Sulfate is dissolved leaving an excess of #H^+#.

Apr 26, 2017

Answer:

Well stated. Due to the relatively high Ka value of HSO4-
, the formation of the bisulfate would be very limited. Initial calculation of conversion of sulfate to bisulfate is ~9.93E-6M in a pH~2 acid solution. So, based on this, I stand corrected as such a small conversion would certainly indicate poor solubility. I submit my quantitative in the adjacent 'add an answer' section.

Explanation:

Wiki reports stomach acid pH ~ 1.5 - 3.5. Using a pH~2 and adding #BaSO_4# the following is the relative solubility ...

#BaSO_4#(s) #rightleftharpoons# #Ba^(+2)#(aq) + #SO_4^(-2)#(aq); #K_(sp)# = #9.85 x 10^(-9)#
#SO_4^(-2)#(aq) + #H^+#(aq)#rightleftharpoons# #HSO_4^(-2)#; #K_f# = #1/K_a# = #1/0.11#=90.91
...........................................................................................................................
#BaSO_4#(s) + #H^+#(aq)#rightleftharpoons##HSO_4^(-2)#(aq) + #Ba^(+2)#(aq); K(net) = #K_(sp)#x#K_f# = #9.85 x 10^(-9)#
If acid solution has pH ~ 2 => #[H^+]# = #0.01M#

for K(net) = #[HSO_4^-][Ba^(+2)]#/#[H^+]# = #X^2/0.010#=#9.85x10^-9# => X = Solubility of #BaSO_4#=9.93 x #10^(-6)#M (Very poor solubility indeed). Thanks for the heads up.