# Question 4c428

Apr 26, 2017

When HCl is added into a $B a S {O}_{4}$(aq) solution, the conjugate base $\left(S {O}_{4}^{-} 2\right)$ adds a proton to => $\left(H S {O}_{4}^{-} 2\right)$ a weak acid and $B a S {O}_{4}$ solubility is increased.

#### Explanation:

${H}_{2} S {O}_{4}$ is one of the 6 strong acids (HCl, HBr, HI, HNO_3, HClO_4, H_2SO_4#). However, for ${H}_{2} S {O}_{4}$; 100% ionization occurs only in the 1st ionization step as the bisulfate is a weak acid $\left(H S {O}_{4}^{-}\right)$ with ${K}_{a}$ = $1.1 X {10}^{-} 2$. 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 $\left(S {O}_{4}^{-} 2\right)$ from the $B a S {O}_{4}$ equilibrium causing more $B a S {O}_{4}$ to dissolve and ionize in order to replace the $\left(S {O}_{4}^{-} 2\right)$ removed from the solution system.

Same Explanation but with equilibrium rxn of $B a S {O}_{4}$
Adding HCl => ${H}^{+}$ into $B a S {O}_{4}$(s) $r i g h t \le f t h a r p \infty n s$ "$B {a}^{+ 2}$ + $S {O}_{4}^{2 -}$
=>$B a S {O}_{4}$(s) $r i g h t \le f t h a r p \infty n s$ "$B {a}^{+ 2}$ + $H S {O}_{4}^{2 -}$ where $H S {O}_{4}^{2 -}$ is no longer a part of the $B a S {O}_{4}$ equilibrium. This causes reaction to shift right to deliver more $S {O}_{4}^{2 -}$ 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

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 $B a S {O}_{4}$ the following is the relative solubility ...

$B a S {O}_{4}$(s) $r i g h t \le f t h a r p \infty n s$ $B {a}^{+ 2}$(aq) + $S {O}_{4}^{- 2}$(aq); ${K}_{s p}$ = $9.85 x {10}^{- 9}$
$S {O}_{4}^{- 2}$(aq) + ${H}^{+}$(aq)$r i g h t \le f t h a r p \infty n s$ $H S {O}_{4}^{- 2}$; ${K}_{f}$ = $\frac{1}{K} _ a$ = $\frac{1}{0.11}$=90.91
...........................................................................................................................
$B a S {O}_{4}$(s) + ${H}^{+}$(aq)$r i g h t \le f t h a r p \infty n s$$H S {O}_{4}^{- 2}$(aq) + $B {a}^{+ 2}$(aq); K(net) = ${K}_{s p}$x${K}_{f}$ = $9.85 x {10}^{- 9}$
If acid solution has pH ~ 2 => $\left[{H}^{+}\right]$ = $0.01 M$

for K(net) = $\left[H S {O}_{4}^{-}\right] \left[B {a}^{+ 2}\right]$/$\left[{H}^{+}\right]$ = ${X}^{2} / 0.010$=$9.85 x {10}^{-} 9$ => X = Solubility of $B a S {O}_{4}$=9.93 x ${10}^{- 6}$M (Very poor solubility indeed). Thanks for the heads up.