250 mL of 0.5 M NaOH was added to 500 mL of 1 M HCl. The number of unreacted HCl molecules in the so

Q: Why do we calculate millimoles instead of moles directly?

Since the volume is given in milliliters (mL), multiplying it directly by molarity (mol/L) gives millimoles (mmol). It saves us from converting volumes to liters early on and avoids dealing with tiny decimal numbers.

Q: How did we know NaOH is the limiting reagent?

The balanced equation $\text{NaOH} + \text{HCl} \longrightarrow \text{NaCl} + \text{H}_2\text{O}$ shows they react in a 1:1 ratio. Since we have 125 mmol of NaOH and 500 mmol of HCl, NaOH is present in a smaller amount and will be completely consumed first.

Q: Why did we multiply by $10^{-3}$ in the final calculation?

The unreacted amount we found was 375 **millimoles**. To use Avogadro's number ($N_A$), which is defined per **mole**, we must convert millimoles back to moles by multiplying by $10^{-3}$.

Q: What if the reaction wasn't 1:1, like with $\text{H}_2\text{SO}_4$?

If it were $\text{H}_2\text{SO}_4$, the reaction would be $2\text{NaOH} + \text{H}_2\text{SO}_4 \longrightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$. You would need 2 moles of NaOH for every 1 mole of acid, which changes the limiting reagent calculation completely!

Q: How do I avoid calculation errors with large powers of 10?

Always group your standard numbers and your powers of 10 separately. Multiply $375 \times 6.022$ first, and then combine $10^{-3} \times 10^{23} = 10^{20}$. Finally, adjust the decimal point to match the requested format ($10^{21}$).

Question

250 mL of 0.5 M NaOH was added to 500 mL of 1 M HCl. The number of unreacted HCl molecules in the solution after complete reaction is $\dots 	imes 10^{21}$. (Nearest integer)
($N_A = 6.022 	imes 10^{23}$)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Let's visualize the setup. We have two solutions ready to be mixed. On one side, we have $250 	ext{ mL}$ of $0.5 	ext{ M}$ sodium hydroxide ($	ext{NaOH}$). On the other side, a much larger and more concentrated batch: $500 	ext{ mL}$ of $1 	ext{ M}$ hydrochloric acid ($	ext{HCl}$). Imagine pouring them together into a single large beaker.

To find out exactly how these chemicals will react, we need to count their particles. Since our volumes are in milliliters, it's incredibly convenient to calculate the number of millimoles. The formula is simple: millimoles equal volume in milliliters multiplied by molarity.

$$n 	ext{ (mmol)} = V 	ext{ (mL)} 	imes M 	ext{ (mol/L)}$$

### Calculating Initial Millimoles

Let's substitute our given values into this formula. For sodium hydroxide, we multiply $250$ by $0.5$. And for hydrochloric acid, we multiply $500$ by $1$.

$$n_{	ext{NaOH}} = 250 	imes 0.5 = 125 	ext{ mmol}$$

$$n_{	ext{HCl}} = 500 	imes 1 = 500 	ext{ mmol}$$

Notice the huge difference in their amounts! We have four times as much acid as we do base.

### The Neutralization Reaction

When we mix a strong base like sodium hydroxide with a strong acid like hydrochloric acid, a classic neutralization reaction occurs. They react in a simple one-to-one ratio to form sodium chloride and water.

$$	ext{NaOH} + 	ext{HCl} \longrightarrow 	ext{NaCl} + 	ext{H}_2	ext{O}$$

Let's set up our stoichiometry table. At time $t=0$, right before the reaction starts, we have $125 	ext{ mmol}$ of base and $500 	ext{ mmol}$ of acid. No products have been formed yet.

Since they react one-to-one, the reactant with the smaller amount will run out first. That's our **limiting reagent**. Here, sodium hydroxide is the limiting reagent. All $125 	ext{ mmol}$ of it will be completely consumed, taking exactly $125 	ext{ mmol}$ of hydrochloric acid down with it.

So, what's left in the beaker after the reaction stops? The base is completely gone. But for the acid, we started with $500$ and used up $125$. Subtracting those gives us $375 	ext{ mmol}$ of unreacted hydrochloric acid floating around.

$$	ext{Unreacted HCl} = 500 - 125 = 375 	ext{ mmol}$$

### Moles to Molecules

The question asks for the number of molecules, not millimoles. First, we convert $375 	ext{ millimoles}$ to moles by multiplying by $10^{-3}$. Then, we multiply by Avogadro's number ($N_A$) to find the total molecule count.

$$	ext{Number of molecules} = 	ext{Moles} 	imes N_A$$

Let's plug in the numbers. We have $375 	imes 10^{-3} 	ext{ moles}$, multiplied by Avogadro's constant, which is given as $6.022 	imes 10^{23}$. Don't make a silly mistake here with the powers of ten.

$$	ext{Molecules} = (375 	imes 10^{-3}) 	imes (6.022 	imes 10^{23})$$

### Final Calculation

Let's group the terms. $375$ multiplied by $6.022$ gives us $2258.25$. And $10^{-3}$ times $10^{23}$ simplifies to $10^{20}$.

$$	ext{Molecules} = 2258.25 	imes 10^{20}$$

Shifting the decimal one place to the left, we get $225.825 	imes 10^{21}$.

The question asks for the value of $x$ to the nearest integer, where the answer is in the format $x 	imes 10^{21}$. Rounding $225.825$ to the nearest whole number gives us exactly **$226$**. And that is our final answer!

Animated Solution for Chemistry - Basic Concepts in Chemistry: 250 mL of 0.5 M NaOH was added to 500 mL of 1 M HCl. The number of unreacted HCl molecules in the solution after complete reaction is $\dots \times 1 0^{21}$ . (Nearest integer) ( $N_{A} = 6.022 \times 1 0^{23}$ )

Enter Numerical Value:

Visualized Solution

\text{Analyzing the Reactants}

\text{Millimoles Formula}

\text{Calculating Initial Millimoles}

\text{Initial Millimoles Values}

\text{The Neutralization Reaction}

\text{Stoichiometry at } t=0

\text{Identifying the Limiting Reagent}

\text{Unreacted HCl}

\text{Moles to Molecules}

\text{Substituting Avogadro's Number}

\text{Final Calculation}

\text{Rounding to Nearest Integer}

\text{Food for Thought}

The Sigma Insight: Stoichiometric and Volumetric Calculations

Analyzing the Setup

Calculating Initial Millimoles

The Neutralization Reaction

Moles to Molecules

Final Calculation

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