Problem 46 - Chemistry Matter and Change

Question

Challenge How many milliliters of 0.500 \(\mathrm{M}\) NaOH would neutralize 25.00 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4} ?\)

Step-by-Step Solution

Answer

15.00 mL of 0.500 M NaOH solution is needed to neutralize 25.00 mL of 0.100 M \(\mathrm{H}_{3} \mathrm{PO}_{4}\) solution.

Step 1: Write the balanced chemical equation

We start by writing the balanced chemical equation for the neutralization reaction between phosphoric acid (\(\mathrm{H}_{3}\mathrm{PO}_{4}\)) and sodium hydroxide (NaOH). The equation for this reaction is: \[ \mathrm{H}_{3} \mathrm{PO}_{4} + 3 \mathrm{NaOH} \rightarrow \mathrm{Na}_{3} \mathrm{PO}_{4} + 3 \mathrm{H}_{2} \mathrm{O} \]

Step 2: Determine moles of phosphoric acid

To find the moles of phosphoric acid, we use the given information of the volume and concentration of the solution: Moles of acid = volume × concentration \[ \text{moles of } \mathrm{H}_{3}\mathrm{PO}_{4} = (25.00 \: \mathrm{mL})(0.100 \: \mathrm{M}) \] Convert mL to L: \[ (25.00 \: \mathrm{mL})(\frac{1 \: \mathrm{L}}{1000 \: \mathrm{mL}}) = 0.025 \: \mathrm{L} \] Now, we can calculate the moles of phosphoric acid: \[ \text{moles of } \mathrm{H}_{3}\mathrm{PO}_{4} = (0.025 \: \mathrm{L})(0.100 \: \mathrm{M}) = 0.0025 \: \mathrm{mol} \]

Step 3: Calculate moles of sodium hydroxide needed

Now, we need to find the moles of sodium hydroxide required to neutralize the given amount of phosphoric acid. From the balanced equation, we know that 1 mole of \(\mathrm{H}_{3}\mathrm{PO}_{4}\) requires 3 moles of NaOH: \[ \text{moles of }\mathrm{NaOH} = \text{(moles of }\mathrm{H}_{3}\mathrm{PO}_{4})(3) = (0.0025 \: \mathrm{mol})(3) = 0.0075 \: \mathrm{mol} \]

Step 4: Calculate the volume of sodium hydroxide solution needed

Using the moles of NaOH and the given concentration of the NaOH solution, we can find the volume of NaOH solution needed to neutralize the phosphoric acid: Moles of NaOH = volume × concentration \[ \text{volume of NaOH} = \frac{\text{moles of NaOH}}{\text{concentration of NaOH}} \] \[ \text{volume of NaOH} = \frac{0.0075 \: \mathrm{mol}}{0.500 \: \mathrm{M}} = 0.015 \: \mathrm{L} \] Finally, convert liters to milliliters: \[ 0.015 \: \mathrm{L} \times \frac{1000 \: \mathrm{mL}}{1 \: \mathrm{L}} = 15.00 \: \mathrm{mL} \] Therefore, 15.00 mL of 0.500 M NaOH solution is needed to neutralize 25.00 mL of 0.100 M \(\mathrm{H}_{3} \mathrm{PO}_{4}\) solution.

Key Concepts

Chemical Equation Balancing

Understanding how to balance chemical equations is a foundational skill in chemistry. It involves making sure that the number of atoms of each element is the same on both sides of the arrow, thereby respecting the law of conservation of mass. Take the neutralization reaction between phosphoric acid (\( \text{H}_3\text{PO}_4 \) and sodium hydroxide (NaOH) as an example. The balanced equation is:

\[\begin{equation}\text{H}_3\text{PO}_4 + 3\text{NaOH} \rightarrow \text{Na}_3\text{PO}_4 + 3\text{H}_2\text{O}\[\begin{equation}
This shows that one mole of phosphoric acid reacts with three moles of sodium hydroxide to produce one mole of sodium phosphate and three moles of water. Balancing this equation was a critical first step in solving the original exercise as it guides us to the proper ratio of reactants required for a complete reaction, which is essential for the subsequent stoichiometric calculations.

Molarity and Concentration

Molarity is a measure of concentration that signifies the number of moles of a solute in one liter of solution. It's represented by the symbol M and is a critical part of the solution calculation process, as seen in our neutralization exercise. When provided with a molarity of 0.100 M for phosphoric acid and 0.500 M for sodium hydroxide, we are being told exactly how concentrated these solutions are. The ability to interpret and utilize molarity allows us to determine the amount of each chemical present in a given volume of solution.

For the problem presented, we multiply the volume of the phosphoric acid solution by its concentration to find out how many moles of acid are in the solution. Similarly, we would do the same with the base once we determine the required volume to neutralize the acid. This step is vital for quantifying the substances involved in the neutralization reaction and feeds into the stoichiometry for the overall process.

Stoichiometry

The term stoichiometry is derived from the Greek words 'stoicheion' (element) and 'metron' (measure), which together mean the measurement of elements. In chemistry, stoichiometry refers to the quantitative relationship between the reactants and products in a balanced chemical equation. In the context of our neutralization exercise, we employed stoichiometry to calculate the amount of sodium hydroxide needed to completely react with the given amount of phosphoric acid.

We first determined the moles of the acid and then used the mole ratio from the balanced chemical equation to find the moles of the base required. The balanced equation shows the stoichiometric ratio of 1:3 for \( \text{H}_3\text{PO}_4 \) to NaOH, which dictates that three times as much NaOH is needed as phosphoric acid in terms of moles. Once the necessary moles of NaOH were calculated, we were able to find out the volume of the NaOH solution required using its molarity. Stoichiometry is essentially a form of chemical bookkeeping that ensures that reactions are accurately described and quantified.