Chapter 6: Problem 72
What volume of each of the following bases will react completely with \(25.00 \mathrm{mL}\) of \(0.200 \mathrm{M}\) HCl? a. \(0.100 M\) NaOH c. \(0.250 M\) KOH b. \(0.0500 M \operatorname{Sr}(O H)_{2}\)
Short Answer
Expert verified
a. To react completely with $25.00 \mathrm{mL}$ of $0.200 \mathrm{M}$ HCl, you will need $50.00 \mathrm{mL}$ of $0.100 \mathrm{M}$ NaOH.
b. To react completely with $25.00 \mathrm{mL}$ of $0.200 \mathrm{M}$ HCl, you will need $20.00 \mathrm{mL}$ of $0.250 \mathrm{M}$ KOH.
c. To react completely with $25.00 \mathrm{mL}$ of $0.200 \mathrm{M}$ HCl, you will need $50.00 \mathrm{mL}$ of $0.0500 \mathrm{M} \operatorname{Sr}(O H)_{2}$.
Step by step solution
01
Write the balanced chemical equation
Write the balanced chemical equation for the reaction between NaOH and HCl:
NaOH + HCl → NaCl + H₂O
The stoichiometric coefficient for NaOH and HCl is 1.
02
Calculate moles of HCl
Find the number of moles of HCl present in 25.00 mL of 0.200 M HCl:
moles of HCl = volume (L) × concentration (mol/L)
moles of HCl = (25.00 mL × 1 L/1000 mL) × 0.200 mol/L
moles of HCl = 0.00500 moles
03
Calculate moles of NaOH
The reaction between NaOH and HCl is in a 1:1 ratio. Therefore, we need the same number of moles of NaOH to completely react with HCl:
moles of NaOH = 0.00500 moles
04
Calculate volume of NaOH
Now, divide the moles of NaOH by its concentration (0.100 M) to get the volume of NaOH needed:
volume of NaOH = moles of NaOH / concentration
volume of NaOH = 0.00500 moles / 0.100 M
volume of NaOH = 0.0500 L = 50.00 mL
#For part (b):#
05
Write the balanced chemical equation
Write the balanced chemical equation for the reaction between KOH and HCl:
KOH + HCl → KCl + H₂O
The stoichiometric coefficient for KOH and HCl is 1.
06
Calculate moles of KOH
The reaction between KOH and HCl is in a 1:1 ratio. Therefore, we need the same number of moles of KOH to completely react with HCl:
moles of KOH = 0.00500 moles
07
Calculate volume of KOH
Now, divide the moles of KOH by its concentration (0.250 M) to get the volume of KOH needed:
volume of KOH = moles of KOH / concentration
volume of KOH = 0.00500 moles / 0.250 M
volume of KOH = 0.0200 L = 20.00 mL
#For part (c):#
08
Write the balanced chemical equation
Write the balanced chemical equation for the reaction between Sr(OH)₂ and HCl:
Sr(OH)₂ + 2HCl → SrCl₂ + 2H₂O
The stoichiometric coefficient for Sr(OH)₂ is 1, and for HCl is 2.
09
Calculate moles of Sr(OH)₂
The reaction between Sr(OH)₂ and HCl is in a 1:2 ratio. Therefore, we need half the number of moles of Sr(OH)₂ to completely react with HCl:
moles of Sr(OH)₂ = 0.00500 moles / 2
moles of Sr(OH)₂ = 0.00250 moles
10
Calculate volume of Sr(OH)₂
Now, divide the moles of Sr(OH)₂ by its concentration (0.0500 M) to get the volume of Sr(OH)₂ needed:
volume of Sr(OH)₂ = moles of Sr(OH)₂ / concentration
volume of Sr(OH)₂ = 0.00250 moles / 0.0500 M
volume of Sr(OH)₂ = 0.0500 L = 50.00 mL
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Stoichiometry
Understanding stoichiometry is essential for mastering acid-base titration calculations. Stoichiometry is the study of the quantitative relationships between the amounts of reactants used and products formed in a chemical reaction, based on the balanced chemical equations. A stoichiometric amount describes the exact proportion of reactants needed to react without any excess.
For instance, the reaction between NaOH and HCl has a stoichiometric coefficient of 1 for both reactants, indicating a 1:1 ratio. In other words, one mole of NaOH will react with one mole of HCl to produce one mole of NaCl and one mole of water. When you know the stoichiometry of a reaction, you can precisely determine the volume of another reactant needed to completely react with a known quantity of one reactant, which is often the goal in titration.
For instance, the reaction between NaOH and HCl has a stoichiometric coefficient of 1 for both reactants, indicating a 1:1 ratio. In other words, one mole of NaOH will react with one mole of HCl to produce one mole of NaCl and one mole of water. When you know the stoichiometry of a reaction, you can precisely determine the volume of another reactant needed to completely react with a known quantity of one reactant, which is often the goal in titration.
Molarity
Molarity is a measure of the concentration of a solute in a solution and is defined as the number of moles of solute per liter of solution. It is denoted as M and is a crucial concept in titration because it allows chemists to relate the volume of a solution to the amount of substance in moles.
The formula to calculate molarity is given by:
\( \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \).
For example, if you have a solution of NaOH with a molarity of 0.100 M, this means each liter of solution contains 0.100 moles of NaOH. By knowing the molarity and the volume of the solution, you can calculate the number of moles present, which is vital for figuring out how much of another substance will react with it in a titration process.
The formula to calculate molarity is given by:
\( \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \).
For example, if you have a solution of NaOH with a molarity of 0.100 M, this means each liter of solution contains 0.100 moles of NaOH. By knowing the molarity and the volume of the solution, you can calculate the number of moles present, which is vital for figuring out how much of another substance will react with it in a titration process.
Balanced Chemical Equations
Balanced chemical equations are representations of chemical reactions where the number of atoms for each element is the same on both the reactant and product sides of the equation. This balance is a requirement because of the law of conservation of mass.
When writing equations for titration reactions, it’s essential that they are balanced in order to correctly perform moles calculation and stoichiometry. Take the example of the reaction between HCl and NaOH forming NaCl and H2O:
\( \text{NaOH + HCl} \rightarrow \text{NaCl + H2O} \).
In this equation, each element has the same number of atoms on both sides, making the equation balanced. The balanced equation tells us that one mole of NaOH reacts with one mole of HCl, which is fundamental information when calculating how much of one reactant is needed to completely react with a given amount of another.
When writing equations for titration reactions, it’s essential that they are balanced in order to correctly perform moles calculation and stoichiometry. Take the example of the reaction between HCl and NaOH forming NaCl and H2O:
\( \text{NaOH + HCl} \rightarrow \text{NaCl + H2O} \).
In this equation, each element has the same number of atoms on both sides, making the equation balanced. The balanced equation tells us that one mole of NaOH reacts with one mole of HCl, which is fundamental information when calculating how much of one reactant is needed to completely react with a given amount of another.
Moles Calculation
The calculation of moles is a fundamental part of chemical reactions and titration. The mole is a standard unit in chemistry that measures the amount of a substance. One mole is defined as exactly 6.02214076×10²³ elementary entities (usually atoms or molecules).
In acid-base titration calculations, you often start by determining the number of moles in a given volume of solution through the formula:
\( \text{moles} = \text{volume in liters} \times \text{molarity} \).
Thus, for a sample containing 25.00 mL of 0.200 M HCl, convert the volume to liters (0.025 L), then multiply by the molarity to find the moles of HCl:
\( \text{moles of HCl} = 0.025 \text{ L} \times 0.200 \text{ M} = 0.00500 \text{ moles} \).
This calculation is crucial because it allows you to determine how much of another reactant you’ll need to complete the reaction, as seen when you use stoichiometry to calculate the volume of NaOH required to react with the HCl.
In acid-base titration calculations, you often start by determining the number of moles in a given volume of solution through the formula:
\( \text{moles} = \text{volume in liters} \times \text{molarity} \).
Thus, for a sample containing 25.00 mL of 0.200 M HCl, convert the volume to liters (0.025 L), then multiply by the molarity to find the moles of HCl:
\( \text{moles of HCl} = 0.025 \text{ L} \times 0.200 \text{ M} = 0.00500 \text{ moles} \).
This calculation is crucial because it allows you to determine how much of another reactant you’ll need to complete the reaction, as seen when you use stoichiometry to calculate the volume of NaOH required to react with the HCl.
