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I want to calculate the minimum amount of solid barium nitrate (g) needed to be added to $\pu{50mL \text{of} 0.05mol/L}$ sodium hydroxide in order to cause precipitation.

My attempt: $$\ce{2NaOH(aq) + Ba(NO3)2(aq) -> 2NaNO3(aq) + Ba(OH)2(s)} $$

The salt that is going to precipitate is going to be barium hydroxide, which has a $\ce{Ksp = 2.55\times10^{-4}}$. Letting the molar solubility be $s$, I now have $s(2s)^2=2.55\times10^{-4}\Rightarrow s=0.039947\ce{mol/L}$.

$\therefore \ce{ In 50 mL}$, we need $\ce{0.001997 mols}$ for precipitation to occur. However, from the stoichiometry, that means we need $\ce{0.004 mols}$ of $\ce{NaOH}$, whereas there is only $\ce{0.0025 mols}$ to begin with.

So my question is where did my logic go wrong, am I supposed to use the molar solubility at all for this question?

For context, this was a multiple-choice question with options $A=0.339\text{g}, B=0.678\text{g}, C=1.38\text{g} \text{ and } D=2.77\text{g}$.

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    $\begingroup$ Do not expect the ratio of concentrations of precipitated ions would be stoichiometric. What matters is the equation for Ksp. [Ba^2+]=Ksp/[OH-]^2 $\endgroup$
    – Poutnik
    Oct 23, 2022 at 3:44
  • $\begingroup$ @Poutnik If I have $\ce{[OH-] = 0.025}$, that gives me $\ce{[Ba^{2+}] = 40.8}$ if I plug it into the Ksp which seems way too big so I'm not sure what exactly you mean by what matters is the equation for Ksp, could you explain a little more? $\endgroup$ Oct 23, 2022 at 4:00
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    $\begingroup$ $\frac{\pu{2.55E-4}}{0.05^2}=0.102$ $\endgroup$
    – Poutnik
    Oct 23, 2022 at 4:06

1 Answer 1

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The mistake you made when calculating molar solubility $s$ is that you assumed $\ce{Ba(OH)2}$ was dissociating in pure water, when the truth is it was dissociating in a solution of $\ce{NaOH}$.

Since $\ce{OH-}$ is a common ion between $\ce{Ba(OH)2}$ and $\ce{NaOH}$, it will reduce the amount of $\ce{Ba(OH)2}$ that will dissolve compared to a solution of pure water.

In this case, the molar solubility of $\ce{Ba(OH)2}$ would be calculated using this equation:

$$K_{sp}=\ce{[Ba^{2+}]\;[OH^{-}]^2 = (s)\;(0.05+2s)^2}=\;2.55\;*\;10^{-4}$$

Solving for $s$:

$$s=\pu{0.02525 mol / L}$$

Finally, mass of $\ce{Ba(NO3)2}$ is calculated by:

$$m_{\ce{Ba(NO3)2}}=s\;VM_{\ce{Ba(NO3)2}}=\left(\pu{0.02525 mol/L}\right)\left(\pu{0.05 L}\right)\left(\pu{262 g/mol}\right)=\pu{0.331 g}$$

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    $\begingroup$ Thank you, I think the value of s in your answer was typed into the calculator incorrectly but otherwise thanks for spotting the error $\endgroup$ Oct 24, 2022 at 1:00
  • $\begingroup$ You do not dissolve Ba(OH)2. There are no extra OH- from such a dissolution. You dissolve Ba(NO3)2. So Ksp(Ba(OH)2)=[Ba(NO3)2]. (0.05)^2. $\endgroup$
    – Poutnik
    Oct 24, 2022 at 7:17

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