Question

check your volume measurement. did you report your data to the correct number of significant figures? concentration of hydrochloric acid (m) 2.11 concentration of sodium hydroxide (m) 2.05 calorimeter constant (j/°c) 13.4 volume of hydrochloric acid (ml) 93.0 temperature of hydrochloric acid (°c) 24.0 volume of sodium hydroxide (ml) temperature of sodium hydroxide (°c) 24.0 final temperature after mixing (°c) 37.8

Explanation:

To determine the volume of sodium hydroxide (\(NaOH\)) for a neutralization reaction with hydrochloric acid (\(HCl\)), we use the principle of stoichiometry (moles of acid = moles of base for a strong acid - strong base reaction, \(HCl + NaOH
ightarrow NaCl + H_2O\)).

Step 1: Recall the formula for molarity

Molarity (\(M\)) is defined as moles of solute per liter of solution:

$$ M = \frac{\text{moles of solute}}{\text{volume of solution (in L)}} $$

Rearranged to solve for moles:

$$ \text{moles} = M \times \text{volume (in L)} $$

Step 2: Calculate moles of \(HCl\)

Given:

• Concentration of \(HCl\) (\(M_{HCl}\)) = \(2.11 \, M\)
• Volume of \(HCl\) (\(V_{HCl}\)) = \(93.0 \, \text{mL} = 0.0930 \, \text{L}\) (since \(1 \, \text{L} = 1000 \, \text{mL}\))

Using the moles formula:

$$ \text{moles of } HCl = M_{HCl} \times V_{HCl} $$

$$ \text{moles of } HCl = 2.11 \, \frac{\text{mol}}{\text{L}} \times 0.0930 \, \text{L} $$

$$ \text{moles of } HCl = 2.11 \times 0.0930 \, \text{mol} $$

$$ \text{moles of } HCl = 0.19623 \, \text{mol} $$

Step 3: Moles of \(NaOH\) (stoichiometry)

For the reaction \(HCl + NaOH
ightarrow NaCl + H_2O\), the mole ratio of \(HCl\) to \(NaOH\) is \(1:1\). Thus:

$$ \text{moles of } NaOH = \text{moles of } HCl = 0.19623 \, \text{mol} $$

Step 4: Calculate volume of \(NaOH\)

Given:

• Concentration of \(NaOH\) (\(M_{NaOH}\)) = \(2.05 \, M\)

Using the rearranged molarity formula to solve for volume (in L):

$$ V_{NaOH} = \frac{\text{moles of } NaOH}{M_{NaOH}} $$

Substitute the values:

$$ V_{NaOH} = \frac{0.19623 \, \text{mol}}{2.05 \, \frac{\text{mol}}{\text{L}}} $$

$$ V_{NaOH} = \frac{0.19623}{2.05} \, \text{L} $$

$$ V_{NaOH} \approx 0.09572 \, \text{L} $$

Step 5: Convert volume to milliliters

Since \(1 \, \text{L} = 1000 \, \text{mL}\):

$$ V_{NaOH} = 0.09572 \, \text{L} \times 1000 \, \frac{\text{mL}}{\text{L}} $$

$$ V_{NaOH} \approx 95.7 \, \text{mL} $$

Step 6: Check significant figures

The given data (e.g., \(93.0 \, \text{mL}\), \(2.11 \, M\), \(2.05 \, M\)) has 3 significant figures. The calculated volume (\(95.7 \, \text{mL}\)) also has 3 significant figures, consistent with the data.

Answer:

The volume of sodium hydroxide is approximately \(\boldsymbol{95.7 \, \text{mL}}\) (or \(96 \, \text{mL}\) if rounding to 2 significant figures, but 3 is more precise here).