An aqueous solution contains an unknown concentration of Ba2+. When 50 mL of a 1 M solution of Na2SO

Q: Why do we subtract 50 mL from 500 mL at the very beginning?

The problem states the *final* volume of the mixture is 500 mL. This mixture was created by adding 50 mL of sodium sulfate to an unknown volume of barium solution. Therefore, by conservation of volume (assuming ideal mixing), the original barium solution must have been 500 mL - 50 mL = 450 mL.

Q: Why does the concentration of sulfate ions change from 1 M to 0.1 M?

When you pour the 50 mL of 1 M sodium sulfate into the other beaker, the total volume of the liquid expands to 500 mL. The same number of sulfate moles are now spread out over a much larger volume, so their concentration (moles per liter) decreases. This is called dilution.

Q: What exactly does 'just begins to precipitate' mean mathematically?

It means the solution has reached the exact point of saturation. At this precise moment, the product of the concentrations of the reacting ions (the Ionic Product) is exactly equal to the Solubility Product Constant ($K_{sp}$). If the ionic product were any lower, it would be unsaturated. If it were any higher, a solid precipitate would actively form.

Q: I found the barium concentration to be $10^{-9}$ M. Why isn't that the final answer?

That is a very common trap! The $10^{-9}$ M you calculated is the concentration of barium ions *after* the two solutions were mixed together into a 500 mL volume. The question asks for the *original* concentration of the barium solution before the mixing occurred, when it was only 450 mL.

Q: Does the addition of sodium sulfate change the total number of moles of barium ions?

No! Until precipitation actually occurs, the total number of moles of barium ions remains constant. Only their concentration changes because the volume of the solvent they are swimming in has increased. This is why we can use the $M_1V_1 = M_2V_2$ formula, which is essentially a statement of conservation of moles.

Question

An aqueous solution contains an unknown concentration of $	ext{Ba}^{2+}$. When $50	ext{ mL}$ of a $1	ext{ M}$ solution of $	ext{Na}_2	ext{SO}_4$ is added, $	ext{BaSO}_4$ just begins to precipitate. The final volume is $500	ext{ mL}$. The solubility product of $	ext{BaSO}_4$ is $1 	imes 10^{-10}$. What is the original concentration of $	ext{Ba}^{2+}$?

Options:

(A) $5 	imes 10^{-9}	ext{ M}$

(B) $2 	imes 10^{-9}	ext{ M}$

(C) $1.1 	imes 10^{-9}	ext{ M}$

(D) $1.0 	imes 10^{-10}	ext{ M}$

Sigma Percentile · Accepted Answer

### The Mystery of the Unknown Concentration

Imagine you are standing in a chemistry laboratory. On your workbench sits a beaker containing a clear, colorless solution of barium ions ($Ba^{2+}$). The label is missing, and you have absolutely no idea what its concentration is. This is the mystery we need to solve!

To find out, you decide to perform a classic precipitation experiment. You take exactly $50	ext{ mL}$ of a $1	ext{ M}$ sodium sulfate ($Na_2SO_4$) solution and pour it into the unknown barium solution. The problem tells us that the final volume of this newly mixed solution is exactly $500	ext{ mL}$.

Before we even think about chemical reactions, let's do some basic accounting of our volumes. If the final mixture is $500	ext{ mL}$ and we added $50	ext{ mL}$ of sodium sulfate, what was the volume of our original, mysterious barium solution?

$$V_{	ext{Ba}^{2+}} = V_{	ext{final}} - V_{	ext{added}} = 500	ext{ mL} - 50	ext{ mL} = 450	ext{ mL}$$

So, our original unknown solution had a volume of $450	ext{ mL}$. Keep this number safe; we will need it later to travel back in time!

### The Dilution of Sulfate Ions

Now, let's shift our focus to the sulfate ions ($SO_4^{2-}$). When we poured that $50	ext{ mL}$ of $1	ext{ M}$ sodium sulfate into the larger beaker, it didn't just sit there. It mixed completely, and its volume expanded to fill the entire $500	ext{ mL}$ space.

Because the volume increased, the concentration of the sulfate ions must have decreased. They got diluted! To find their new concentration in the final mixture, we rely on the fundamental dilution equation:

$$M_1V_1 = M_2V_2$$

Let's plug in the values for our sulfate solution:

$$1	ext{ M} 	imes 50	ext{ mL} = M_{	ext{SO}_4^{2-}} 	imes 500	ext{ mL}$$

Solving for the new molarity ($M_{	ext{SO}_4^{2-}}$):

$$M_{	ext{SO}_4^{2-}} = \frac{1 	imes 50}{500} = \frac{50}{500} = 0.1	ext{ M}$$

This $0.1	ext{ M}$ is the concentration of sulfate ions floating around in our final $500	ext{ mL}$ mixture, ready to react.

### The Tipping Point: Just Precipitation

Here is the most critical phrase in the entire problem: **"$BaSO_4$ just begins to precipitate."**

What does this mean physically? It means the solution has reached its absolute limit of holding dissolved barium and sulfate ions. It is perfectly saturated. Mathematically, this "tipping point" occurs exactly when the **Ionic Product** equals the **Solubility Product Constant ($K_{sp}$)**.

$$[	ext{Ba}^{2+}][	ext{SO}_4^{2-}] = K_{sp}$$

We are given that the $K_{sp}$ for barium sulfate is $1 	imes 10^{-10}$. We also just calculated that the sulfate concentration in the mixture is $0.1	ext{ M}$. Let's substitute these values into our equilibrium condition:

$$[	ext{Ba}^{2+}] 	imes 0.1 = 1 	imes 10^{-10}$$

Now, we can easily isolate the concentration of barium ions present in the final mixture:

$$[	ext{Ba}^{2+}] = \frac{1 	imes 10^{-10}}{0.1} = 10^{-9}	ext{ M}$$

**Warning!** Do not fall into the trap of selecting this as your final answer. This $10^{-9}	ext{ M}$ is the concentration of barium ions in the *final, diluted $500	ext{ mL}$ mixture*, not in the original beaker!

### Working Backwards to the Source

To find the original concentration of the barium solution, we have to reverse the dilution process. Before mixing, all those barium ions were confined to a smaller volume of $450	ext{ mL}$. When the volume expanded to $500	ext{ mL}$, their concentration dropped to $10^{-9}	ext{ M}$.

We apply our trusty dilution formula one last time, but now for the barium ions:

$$M_{	ext{initial}} 	imes V_{	ext{initial}} = M_{	ext{final}} 	imes V_{	ext{final}}$$

$$M_{	ext{initial}} 	imes 450	ext{ mL} = 10^{-9}	ext{ M} 	imes 500	ext{ mL}$$

Let's solve for the initial molarity:

$$M_{	ext{initial}} = \frac{500}{450} 	imes 10^{-9}$$

$$M_{	ext{initial}} = \frac{10}{9} 	imes 10^{-9}$$

$$M_{	ext{initial}} \approx 1.11 	imes 10^{-9}	ext{ M}$$

And there we have it! The original concentration of our mysterious barium solution was approximately $1.11 	imes 10^{-9}	ext{ M}$. The mystery is solved, and the beauty of chemical equilibrium is revealed once again.

Select Answer:

Visualized Solution

$V_{initial} of Ba^{2 +}$

$Concentration of SO_{4}^{2 -} in mixture$

$M_{SO_{4}^{2 -}}$

$Condition for Precipitation$

$[Ba^{2 +}] in mixture$

$[Ba^{2 +}] in mixture$

$Original [Ba^{2 +}]$

$Final Calculation$

$What if?$

The Sigma Insight: Solubility Product and Common Ion Effect

The Mystery of the Unknown Concentration

The Dilution of Sulfate Ions

The Tipping Point

Working Backwards to the Source

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vinitial​ of Ba2+

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Concentration of SO42−​ in mixture

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } MSO42−​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Precipitation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [Ba2+] in mixture

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [Ba2+] in mixture

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Original [Ba2+]

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if?

The Sigma Insight: Solubility Product and Common Ion Effect

The Mystery of the Unknown Concentration

The Dilution of Sulfate Ions

The Tipping Point

Working Backwards to the Source

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