# How do I find the minimum amount needed for precipitation?

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 $$A=0.339\text{g}, B=0.678\text{g}, C=1.38\text{g} \text{ and } D=2.77\text{g}$$.

• 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 Commented Oct 23, 2022 at 3:44
• @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? Commented Oct 23, 2022 at 4:00
• $\frac{\pu{2.55E-4}}{0.05^2}=0.102$ Commented Oct 23, 2022 at 4:06

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:

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

• Thank you, I think the value of s in your answer was typed into the calculator incorrectly but otherwise thanks for spotting the error Commented Oct 24, 2022 at 1:00
• 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. Commented Oct 24, 2022 at 7:17