What is the solubility of "barium sulfate" in a 250*mL volume of 0.01*mol*L^-1 "sulfuric acid"?

K_"sp"("barium sulfate")=1.0842xx10^-10

Apr 20, 2017

We need some data................

Explanation:

This site quotes ${K}_{\text{sp}}$ for $B a S {O}_{4}$ as 1.0842xx10^(−10) under standard conditions.......so it is pretty insoluble stuff.

What does this mean? We it means that if the ion product $\left[B {a}^{2 +}\right] \left[S {O}_{4}^{2 -}\right] = 1.0842 \times {10}^{-} 10$, then equilibrium exists between the species in solution, and the solid salt.

If $\left[B {a}^{2 +}\right] \left[S {O}_{4}^{2 -}\right] > 1.0842 \times {10}^{-} 10$, then precipitation of $B a S {O}_{4}$ will occur until equality is reached.

So, I think you have specified that $S {O}_{4}^{2 -} = 0.01 \cdot m o l \cdot {L}^{-} 1$. And thus, at saturation point, if we add $B {a}^{2 +}$ ion then..........

$\left[B {a}^{2 +}\right] \equiv \frac{1.0842 \times {10}^{-} 10}{\left[S {O}_{4}^{2 -}\right]}$

$\equiv \frac{1.0842 \times {10}^{-} 10}{0.01 \cdot m o l \cdot {L}^{-} 1} = 1.08 \times {10}^{-} 8 \cdot m o l \cdot {L}^{-} 1$

And so in a $250 \cdot m L$ volume, then we have a mass of...........

$1.08 \times {10}^{-} 8 \cdot m o l \cdot {L}^{-} 1 \times 0.250 \cdot L \times 208.23 \cdot g \cdot m o {l}^{-} 1$ with respect to $B a C {l}_{2}$, i.e. $=$ $\text{something less than a microgram}$

In pure water, would barium sulfate be more or less soluble? In fact, I think it would be a good idea to quote the solubility of $B a S {O}_{4}$ in $m o l \cdot {L}^{-} 1$ with respect to pure water. Good luck.