Question

Question #4c428

Answer 1

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^+`.

Answer 2

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.