Problem 196
Question
How many moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) would it take to completely neutralize \(0.4\) mole of phosphoric acid, \(\mathrm{H}_{3} \mathrm{PO}_{4} ?\)
Step-by-Step Solution
Verified Answer
To completely neutralize 0.4 mole of phosphoric acid (H3PO4), it would take 0.8 moles of calcium hydroxide (Ca(OH)2).
1Step 1: Write the neutralization equation for the given acid and base
The reaction between phosphoric acid and calcium hydroxide is given by:
\[2\mathrm{Ca(OH)_{2}} + \mathrm{H_{3}PO_{4}} \rightarrow \mathrm{Ca_{2}PO_{4}} + 2\mathrm{H_{2}O}\]
The balanced equation says that 2 moles of calcium hydroxide react with 1 mole of phosphoric acid to give calcium phosphate and water.
2Step 2: Calculate the number of moles of calcium hydroxide
Since one mole of phosphoric acid reacts with two moles of calcium hydroxide, we can use a ratio to calculate how many moles of calcium hydroxide are needed to react with 0.4 mole of phosphoric acid.
If 1 mole of phosphoric acid reacts with 2 moles of calcium hydroxide,
then 0.4 mole of phosphoric acid will react with \( X \) moles of calcium hydroxide.
Setting up the ratio, we have:
\( \frac{1}{0.4} = \frac{2}{X} \)
Solving for \( X \) gives:
\( X = 0.4 \times 2 = 0.8 \)
So, 0.8 moles of calcium hydroxide are required to completely neutralize 0.4 mole of phosphoric acid.
3Step 3: Check the answer
This answer is logical since phosphoric acid has 3 hydrogens and calcium hydroxide only has 2 hydroxide ions. So it takes more moles of the calcium hydroxide to neutralize the phosphoric acid. Therefore, 0.8 moles of calcium hydroxide are necessary to neutralize 0.4 mole of phosphoric acid.
Key Concepts
Mole CalculationBalanced Chemical EquationsAcid-Base Neutralization
Mole Calculation
Mole calculation is essential in the field of chemistry for quantifying substances. One mole corresponds to Avogadro's number, which is approximately 6.022 x 10^23, representing the number of particles (usually atoms or molecules) in that sample. In the given exercise, we utilize mole calculation to determine how many moles of calcium hydroxide (\( \text{Ca(OH)}_2 \) ) are needed to neutralize a specific amount of phosphoric acid (\( \text{H}_3\text{PO}_4 \) ).
It's important to understand that the coefficients in a balanced chemical equation provide the mole ratios between reactants and products. Here, for every one mole of phosphoric acid, two moles of calcium hydroxide are required. When we're given 0.4 moles of phosphoric acid, we set up a simple proportional equation to find that 0.8 moles of calcium hydroxide are needed for complete neutralization.
It's important to understand that the coefficients in a balanced chemical equation provide the mole ratios between reactants and products. Here, for every one mole of phosphoric acid, two moles of calcium hydroxide are required. When we're given 0.4 moles of phosphoric acid, we set up a simple proportional equation to find that 0.8 moles of calcium hydroxide are needed for complete neutralization.
Balanced Chemical Equations
Balanced chemical equations are the bedrock of stoichiometric calculations. They ensure that the Law of Conservation of Mass is satisfied, meaning that the number of atoms for each element is the same on both sides of the equation. The balanced equation in the exercise shows the reaction between phosphoric acid and calcium hydroxide.
The coefficients in front of the chemical formulas are like the recipe for the reaction; they tell you how much of each substance you need or will produce. In our example, the equation states that two moles of calcium hydroxide react with one mole of phosphoric acid to produce calcium phosphate and water. Without balancing, we cannot accurately determine how much of each reactant is needed or how much product is created.
The coefficients in front of the chemical formulas are like the recipe for the reaction; they tell you how much of each substance you need or will produce. In our example, the equation states that two moles of calcium hydroxide react with one mole of phosphoric acid to produce calcium phosphate and water. Without balancing, we cannot accurately determine how much of each reactant is needed or how much product is created.
Acid-Base Neutralization
Acid-base neutralization is a type of chemical reaction where an acid and a base react to form a salt and usually water. This process is crucial in various scientific and industrial processes. In the context of this exercise, phosphoric acid (\( \text{H}_3\text{PO}_4 \) ) is the acid, and calcium hydroxide (\( \text{Ca(OH)}_2 \) ) is the base. Upon reacting, they produce calcium phosphate and water, a typical outcome of a neutralization reaction.
The goal is often to determine the correct amounts of an acid and a base to mix so that they completely neutralize each other. This exercise demonstrated the fundamental stoichiometric concept that more moles of calcium hydroxide are required than moles of phosphoric acid because the acid has three hydrogen ions available for reaction, as opposed to the two hydroxide ions provided by the base.
The goal is often to determine the correct amounts of an acid and a base to mix so that they completely neutralize each other. This exercise demonstrated the fundamental stoichiometric concept that more moles of calcium hydroxide are required than moles of phosphoric acid because the acid has three hydrogen ions available for reaction, as opposed to the two hydroxide ions provided by the base.
Other exercises in this chapter
Problem 194
Other than water, what would you expect to find in the highest concentration in an aqueous solution of KOH? Explain.
View solution Problem 195
What is the \(\mathrm{pH}\) of these aqueous solutions? (a) \(1.0 \mathrm{M} \mathrm{HCl}\), (b) \(0.1 \mathrm{M} \mathrm{HCl}\), (c) \(0.001 \mathrm{M} \mathrm
View solution Problem 198
Sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4}\), is a diprotic acid with dissociation equilibrium constants of \(K_{\mathrm{eq}}>1.0 \times 10^{3}\) and \(K_{
View solution Problem 199
What is the pH of a \(0.010 \mathrm{M}\) aqueous solution of \(\mathrm{NaOH}\) ?
View solution