Question

The electrolyte in a lead storage battery must have a concentration between 4.8 and \(5.3 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) if the battery is to be most effective. A \(5.00 \mathrm{mL}\) sample of a battery acid requires \(49.74 \mathrm{mL}\) of \(0.935 \mathrm{M} \mathrm{NaOH}\) for its complete reaction (neutralization). Does the concentration of the battery acid fall within the desired range? [Hint: Keep in mind that the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) produces two \(\mathrm{H}^{+}\) ions per formula unit.]

Short answer

The concentration of the battery acid (4.66 M) does not fall within the desired range of 4.8-5.3 M for the battery to be most effective.

Step-by-step solution

Determine moles of Sodium Hydroxide

Based on the volume and molarity of sodium hydroxide in the reaction, calculate the number of moles by using the formula: Molarity = Moles / Volume. Rearranging this formula to find moles gives: Moles = Molarity * Volume. Remember to convert volume from mL to L. So, moles of \(\mathrm{NaOH} = 0.935 \mathrm{M}*\frac{49.74 \mathrm{mL}}{1000 \mathrm{mL/L}}=0.0465 \mathrm{mol}\)

Determine moles of Sulfuric Acid

Knowing that every mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) reacts with two moles of \(\mathrm{NaOH}\), we can calculate the moles of sulfuric acid. So, moles of \(\mathrm{H}_{2} \mathrm{SO}_{4} = \frac{0.0465 \mathrm{mol}}{2} = 0.0233 \mathrm{mol}\)

Calculate concentration of Sulfuric Acid

The molarity (concentration in mol/L) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is calculated using the formula: Molarity = Moles / Volume. Volume of acid sample is 5.00 mL, so convert this to litres: Volume = \(\frac{5.00 \mathrm{mL}}{1000 \mathrm{mL/L}}\). Hence, the molarity of sulfuric acid: \(\mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4} = \frac{0.0233 \mathrm{mol}}{0.005 \mathrm{L}} = 4.66 \mathrm{M}\)

Compare Concentration with Desired Range

The molarity found is now compared to the effective range given in the problem (4.8 - 5.3M). Since 4.66 lies outside this range, it can be concluded that the concentration of the battery acid does not fall within the desired range.

Key concepts

Neutralization Reaction

A neutralization reaction occurs when an acid and a base react to form water and a salt. In the context of the lead storage battery, this involves sulfuric acid (\(\mathrm{H}_{2} \mathrm{SO}_{4}\)) reacting with sodium hydroxide (\(\mathrm{NaOH}\)). The chemical equation representing this reaction is:\[\mathrm{H}_{2} \mathrm{SO}_{4} + 2 \mathrm{NaOH} \rightarrow \mathrm{Na}_2\mathrm{SO}_{4} + 2 \mathrm{H}_{2} \mathrm{O}\]Here’s the breakdown:- One molecule of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) provides two hydrogen ions (\(\mathrm{H}^+\)).- Two molecules of \(\mathrm{NaOH}\) provide two hydroxide ions (\(\mathrm{OH}^-\)).- When these combine, they form two water molecules and one molecule of sodium sulfate (\(\mathrm{Na}_2\mathrm{SO}_{4}\)).This reaction is essential in determining the concentration of solutions as involved particles react in whole number ratios. Recognizing the stoichiometry—the way these molecules relate quantitatively—is crucial to understanding how neutralization reactions are calculated and balanced.

Concentration of Solutions

The concentration of a solution is measured in terms of molarity (M), which is the number of moles of solute per liter of solution. For accurate calculations, it’s important to convert volumes from milliliters (mL) to liters (L) by dividing by 1000.Key steps include:- **Calculate moles of solute:** Use the formula \(\text{Moles} = \text{Molarity} \times \text{Volume in L}\). In the given problem, for 49.74 mL of 0.935 M \(\mathrm{NaOH}\), the moles would be: \[ \mathrm{Moles} = 0.935 \times \frac{49.74}{1000} = 0.0465 \]- **Use stoichiometry:** As demonstrated, each \(\mathrm{H}_{2} \mathrm{SO}_{4}\) reacts with two \(\mathrm{NaOH}\) ions.- **Determine concentration:** Calculate the molarity after finding the moles and use the sample volume (converted to L) to get the concentration. In this example, converting the 5.00 mL sample to liters gives you: \[ \text{Volume} = \frac{5.00 \text{ mL}}{1000} = 0.005 \text{ L} \] And therefore, the molarity of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) calculated is \(\frac{0.0233}{0.005} = 4.66 \mathrm{M}\).

Lead Storage Battery Electrolyte

Lead-acid batteries utilize sulfuric acid (\(\mathrm{H}_{2} \mathrm{SO}_{4}\)) as an electrolyte to generate power through electrochemical reactions. For optimal battery performance, the acid concentration should precisely fall between 4.8 M and 5.3 M. A few points regarding the electrolyte's role include:- **Function in battery:** The sulfuric acid facilitates the conduction of ions between the battery’s plates, crucial for maintaining charge and discharge cycles.- **Importance of concentration:** Too low a concentration, like the 4.66 M derived in the example, causes poor efficiency and reduced power.- **Hydrogen ion release:** Each formula unit of the acid releases two hydrogen ions (\(\mathrm{H}^+\)), necessary for conducting electricity within the battery.Maintaining the specified concentration range ensures the battery functions optimally, improving both lifespan and energy capacity of the cell. Thus, monitoring and adjusting acid concentration is vital in battery maintenance and performance.