Question #4c428

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Question #4c428

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Answer 1 · 2017-04-26T13:26:29

When HCl is added into a $B a S O_{4}$ $BaSO_4$ (aq) solution, the conjugate base $(S O_{4}^{- 2})$ $(SO_4^-2)$ adds a proton to => $(H S O_{4}^{- 2})$ $(HSO_4^-2)$ a weak acid and $B a S O_{4}$ $BaSO_4$ solubility is increased.

Explanation:

$H_{2} S O_{4}$ $H_2SO_4$ is one of the 6 strong acids $(H C l, H B r, H I, H N O_{3}, H C l O_{4}, H_{2} S O_{4}$ $(HCl, HBr, HI, HNO_3, HClO_4, H_2SO_4$ ). However, for $H_{2} S O_{4}$ $H_2SO_4$ ; 100% ionization occurs only in the 1st ionization step as the bisulfate is a weak acid $(H S O_{4}^{-})$ $(HSO_4^(-))$ with $K_{a}$ $K_a$ = $1.1 X 10^{- 2}$ $1.1X10^-2$ . 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 $(S O_{4}^{- 2})$ $(SO_4^-2)$ from the $B a S O_{4}$ $BaSO_4$ equilibrium causing more $B a S O_{4}$ $BaSO_4$ to dissolve and ionize in order to replace the $(S O_{4}^{- 2})$ $(SO_4^-2)$ removed from the solution system.

Same Explanation but with equilibrium rxn of $B a S O_{4}$ $BaSO_4$
Adding HCl => $H^{+}$ $H^+$ into $B a S O_{4}$ $BaSO_4$ (s) $⇌$ $rightleftharpoons$ " $B a^{+ 2}$ $Ba^(+2)$ + $S O_{4}^{2 -}$ $SO_4^(2-)$
=> $B a S O_{4}$ $BaSO_4$ (s) $⇌$ $rightleftharpoons$ " $B a^{+ 2}$ $Ba^(+2)$ + $H S O_{4}^{2 -}$ $HSO_4^(2-)$ where $H S O_{4}^{2 -}$ $HSO_4^(2-)$ is no longer a part of the $B a S O_{4}$ $BaSO_4$ equilibrium. This causes reaction to shift right to deliver more $S O_{4}^{2 -}$ $SO_4^(2-)$ 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 $H^{+}$ $H^+$ .

Answer 2 · 2017-04-26T23:47:10

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 $B a S O_{4}$ $BaSO_4$ the following is the relative solubility ...

$B a S O_{4}$ $BaSO_4$ (s) $⇌$ $rightleftharpoons$ $B a^{+ 2}$ $Ba^(+2)$ (aq) + $S O_{4}^{- 2}$ $SO_4^(-2)$ (aq); $K_{s p}$ $K_(sp)$ = $9.85 x 10^{- 9}$ $9.85 x 10^(-9)$
$S O_{4}^{- 2}$ $SO_4^(-2)$ (aq) + $H^{+}$ $H^+$ (aq) $⇌$ $rightleftharpoons$ $H S O_{4}^{- 2}$ $HSO_4^(-2)$ ; $K_{f}$ $K_f$ = $\frac{1}{K_{a}}$ $1/K_a$ = $\frac{1}{0.11}$ $1/0.11$ =90.91
...........................................................................................................................
$B a S O_{4}$ $BaSO_4$ (s) + $H^{+}$ $H^+$ (aq) $⇌$ $rightleftharpoons$ $H S O_{4}^{- 2}$ $HSO_4^(-2)$ (aq) + $B a^{+ 2}$ $Ba^(+2)$ (aq); K(net) = $K_{s p}$ $K_(sp)$ x $K_{f}$ $K_f$ = $9.85 x 10^{- 9}$ $9.85 x 10^(-9)$
If acid solution has pH ~ 2 => $[H^{+}]$ $[H^+]$ = $0.01 M$ $0.01M$

for K(net) = $[H S O_{4}^{-}] [B a^{+ 2}]$ $[HSO_4^-][Ba^(+2)]$ / $[H^{+}]$ $[H^+]$ = $\frac{X^{2}}{0.010}$ $X^2/0.010$ = $9.85 x 10^{- 9}$ $9.85x10^-9$ => X = Solubility of $B a S O_{4}$ $BaSO_4$ =9.93 x $10^{- 6}$ $10^(-6)$ M (Very poor solubility indeed). Thanks for the heads up.

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