Explanations 
Textbooks 
Exams 
GCSE 
IGCSE 
AS 
A Level 
International A Level 
University Admissions Tests 
Flashcards 
Vaia AI 
Notes 
Study Plans 
Study Sets 
Find a job 
Find a degree 
Student Deals 
Magazine 
Mobile App Chapter 15: Problem 58
What volume (mL) of \(0.4233 M \mathrm{H}_{2} \mathrm{SO}_{4}\) is needed to neutralize \(6.38 \mathrm{~g} \mathrm{KOH}\) ?
Short Answer
Expert verified
134.2 mL
Step by step solution
01
Calculate the Moles of KOH
First, we need to find the number of moles of KOH. Use the molar mass of KOH, which is the sum of the atomic masses of K (39.1 g/mol), O (16.0 g/mol), and H (1.0 g/mol). Therefore, the molar mass of KOH is 56.1 g/mol. Now use the formula: \[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \] \[ \text{moles of KOH} = \frac{6.38 \text{ g}}{56.1 \text{ g/mol}} \] \[ \text{moles of KOH} \thickapprox 0.1136 \text{ mol} \]
02
Write the Balanced Chemical Equation
The balanced chemical equation for the neutralization reaction is: \[ \text{H}_{2}\text{SO}_{4} + 2 \text{KOH} \rightarrow \text{K}_{2}\text{SO}_{4} + 2 \text{H}_{2}\text{O} \] This indicates that 1 mole of \(\text{H}_{2}\text{SO}_{4}\) neutralizes 2 moles of \(\text{KOH}\).
03
Calculate the Moles of H₂SO₄ Needed
Using the stoichiometry from the balanced equation, we need to find how many moles of \(\text{H}_{2}\text{SO}_{4}\) are required to neutralize the \(0.1136\) moles of \(\text{KOH}\). Since 1 mole of \(\text{H}_{2}\text{SO}_{4}\) reacts with 2 moles of \(\text{KOH}\): \[ \text{moles of H}_2\text{SO}_4 = \frac{0.1136 \text{ mol KOH}}{2} \] \[ \text{moles of H}_2\text{SO}_4 \thickapprox 0.0568 \text{ mol} \]
04
Calculate the Volume of H₂SO₄ Solution Needed
Finally, to find the volume of \(\text{H}_{2}\text{SO}_{4}\) solution needed, use the molarity formula: \[ \text{Molarity} = \frac{\text{moles}}{\text{volume}} \] Rearrange this to solve for volume: \[ \text{volume} = \frac{\text{moles}}{\text{Molarity}} \] Given that the molarity of \(\text{H}_{2}\text{SO}_{4}\) is \(0.4233\) M, we have: \[ \text{volume of H}_2\text{SO}_4 = \frac{0.0568 \text{ mol}}{0.4233 \text{ M}} \] \[ \text{volume of H}_2\text{SO}_4 \thickapprox 0.1342 \text{ L} \] Convert this to mL: \[ 0.1342 \text{ L} \times 1000 \text{ mL/L} \thickapprox 134.2 \text{ mL} \]
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Stoichiometry
Stoichiometry involves using the relationships in a balanced chemical equation to calculate how much of each substance is involved in a reaction. In our problem, stoichiometry helps us determine how many moles of \(\text{H}_2\text{SO}_4\) are needed to neutralize a given number of moles of \(\text{KOH}\). By referring to the balanced chemical equation, we see the relationship: \(\text{H}_2\text{SO}_4 + 2 \text{KOH} \rightarrow \text{K}_2\text{SO}_4 + 2 \text{H}_2\text{O}\). This tells us that 1 mole of sulfuric acid reacts with 2 moles of potassium hydroxide. By dividing the moles of \(\text{KOH}\) by 2, we get the moles of \(\text{H}_2\text{SO}_4\) needed.
Molar Mass Calculation
Calculating the molar mass of a compound is a foundational step in chemistry calculations. The molar mass is the sum of the atomic masses of all atoms in a molecule. For KOH (potassium hydroxide), we add the atomic masses of potassium (K), oxygen (O), and hydrogen (H). Using the periodic table, we find: K = 39.1 g/mol, O = 16.0 g/mol, H = 1.0 g/mol. Summing these gives us the molar mass of KOH as 56.1 g/mol. This value allows us to convert between the mass of a substance and the number of moles, as shown in the solution where we calculate the moles of \(\text{KOH}\) using \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \).
Volume Calculation
Calculating volume in a molarity problem involves using the formula \( \text{Molarity} = \frac{\text{moles}}{\text{volume}} \). To find the volume needed, we rearrange the formula to \( \text{volume} = \frac{\text{moles}}{\text{Molarity}} \). Given the moles of \(\text{H}_2\text{SO}_4\) and its molarity (M), we substitute these values into the equation. In our example, the moles of sulfuric acid needed are approximately 0.0568 mol, and the molarity is 0.4233 M. The calculated volume of the \(\text{H}_2\text{SO}_4\) solution is approximately 0.1342 L or 134.2 mL after converting liters to milliliters (1 L = 1000 mL).
Balanced Chemical Equation
A balanced chemical equation is crucial because it accurately represents the conservation of mass and the relationships between reactants and products. For our reaction, the balanced equation is \( \text{H}_2\text{SO}_4 + 2 \text{KOH} \rightarrow \text{K}_2\text{SO}_4 + 2 \text{H}_2\text{O} \). This indicates that one mole of sulfuric acid reacts with two moles of potassium hydroxide. Balancing equations ensures that atom counts are the same on both sides, which is critical for performing stoichiometric calculations accurately. Without a balanced equation, we would not get the correct mole ratios needed for further calculations.
