Question

What volume of 0.0521\(M \mathrm{Ba}(\mathrm{OH})_{2}\) is required to neutralize exactly 14.20 \(\mathrm{mL}\) of 0.141 \(\mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4} ?\) Phosphoric acid contains three acidic hydrogens.

Short answer

57.6 mL of 0.0521 M \(\mathrm{Ba}(\mathrm{OH})_{2}\) is required to neutralize exactly 14.20 mL of 0.141 M \(\mathrm{H}_{3} \mathrm{PO}_{4}\).

Step-by-step solution

Find moles of \(\mathrm{H}_{3} \mathrm{PO}_{4}\)

To find the moles of \(\mathrm{H}_{3} \mathrm{PO}_{4}\), use the formula: moles = molarity × volume Here, the molarity is 0.141 M, and the volume is 14.20 mL (which should be converted to Liters to match the units). moles of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) = \(0.141\,\text{M} \times 0.01420\,\text{L}\) = 0.0020022 moles

Find moles of \(\mathrm{Ba}(\mathrm{OH})_{2}\) needed

Since phosphoric acid has three acidic hydrogens and barium hydroxide has two hydroxide ions, the balanced neutralization reaction can be written as: \(2 \mathrm{H}_{3} \mathrm{PO}_{4} + 3 \mathrm{Ba}(\mathrm{OH})_{2} \rightarrow 6 \mathrm{H}_{2}\mathrm{O} + 2 \mathrm{Ba}_{3} (\mathrm{PO}_4)_2\) From the stoichiometry of the balanced equation, 2 moles of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) require 3 moles of \(\mathrm{Ba}(\mathrm{OH})_{2}\). So, moles of \(\mathrm{Ba}(\mathrm{OH})_{2}\) can be determined as follows: moles of \(\mathrm{Ba}(\mathrm{OH})_{2}\) = \(\frac{3}{2}\times\) moles of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) moles of \(\mathrm{Ba}(\mathrm{OH})_{2}\) = \(\frac{3}{2}\times 0.0020022 = 0.0030033\,\text{moles}\)

Find the volume of \(\mathrm{Ba}(\mathrm{OH})_{2}\) needed

Now we know the moles of \(\mathrm{Ba}(\mathrm{OH})_{2}\) required, we can use the molarity of the \(\mathrm{Ba}(\mathrm{OH})_{2}\) solution to determine the volume needed to neutralize the \(\mathrm{H}_{3} \mathrm{PO}_{4}\) solution. We will use the formula: volume = \(\frac{\text{moles}}{\text{molarity}}\) volume of \(\mathrm{Ba}(\mathrm{OH})_{2}\) needed = \(\frac{0.0030033\,\text{moles}}{0.0521\,\text{M}}\) = 0.0576 L To convert the volume to milliliters, multiply by 1000: volume of \(\mathrm{Ba}(\mathrm{OH})_{2}\) needed = 57.6 mL So, 57.6 mL of 0.0521 M \(\mathrm{Ba}(\mathrm{OH})_{2}\) is required to neutralize exactly 14.20 mL of 0.141 M \(\mathrm{H}_{3} \mathrm{PO}_{4}\).

Key concepts

Acid-Base Neutralization

In chemistry, acid-base neutralization is a reaction where an acid reacts with a base to produce water and a salt. It’s a fundamental type of chemical reaction. In the given problem, phosphoric acid ( H_3PO_4 ) reacts with barium hydroxide ( Ba(OH)_2 ). This process can be understood better by knowing that phosphoric acid has three hydrogen ions to donate, making it a triprotic acid. Each molecule can neutralize up to three hydroxide ions.

• For every mole of phosphoric acid, we need a specific amount of the base for complete neutralization.
• Barium hydroxide dissociates in solution to provide two hydroxide ions per formula unit, engaging in the neutralization process to form water ( H_2O ).

Overall, the complete reaction forms barium phosphate as a precipitate, along with water as a product.

Molarity Calculations

Molarity is a way to express the concentration of a solution, defined as the number of moles of solute per liter of solution. In this problem, you start by calculating the moles of H_3PO_4. Begin with its given molarity (0.141 M) and volume (14.20 mL), converting milliliters to liters (0.01420 L). Multiplying these gives the total moles available in the solution:\[moles\,of\,H_3PO_4 = 0.141\,M \times 0.01420\,L = 0.0020022\,\text{moles}\]

• The next step involves relating these moles to those of Ba(OH)_2 using the stoichiometric relationship from the balanced chemical equation.
• Convert the moles of hydroxide needed into corresponding volume, using its molarity to ensure precision in neutralization.

Ultimately, accurate molarity calculations ensure that supplies are adequate for conducting the entire reaction without errors.

Chemical Reaction Balancing

Balancing chemical equations involves equalizing the number of each type of atom on both sides of the equation. This ensures the law of conservation of mass is upheld, meaning matter is not created or destroyed in a chemical reaction. In the provided example:\[2 \;(H_3PO_4) + 3 \;(Ba(OH)_2) \rightarrow 6 \;(H_2O) + 1 \;(Ba_3(PO_4)_2)\]

• The balanced equation reveals that 2 moles of phosphoric acid react with 3 moles of barium hydroxide.
• Each part of the equation is balanced by adjusting coefficients, ensuring each element has the same quantity on both sides of the arrow.

Once balanced, this equation can guide the stoichiometry calculations needed to find precisely how much of each reactant is necessary for full neutralization, optimizing the resources and minimizing waste.