The concentration of 20.0 mL H2SO4 solution that requires 22.87 mL of a 0.158 M KOH solution to reach equivalence point is 0.0904 M. or 0.0904 moles/L
Further Explanation:
Molarity
- Molarity may be defined as the concentration of a solution in moles per liter. It is calculated by dividing the number of moles of a solution, n by its volume in liters, V.
- That is; Molarity = n/V, where n is the number of moles and V is the volume in liters.
- Therefore; when given the molarity of a solution and the volume of the solution, then we can get the number of moles, by multiplying the molarity by volume;
- Number of moles = Molarity x Volume
- Additionally, the number of moles of a compound or an element can also be calculated by dividing the mass of the compound by its relative molecular mass.
In the question given;
The equation for the reaction is:
H2SO4 + 2KOH = K2SO4 + 2H2O
Step 1: Moles of KOH
Number of moles = Molarity x Volume
Molarity of KOH is 0.158 M
Volume of KOH is 22.87 mL or 0.02287 L
Therefore;
Number of moles = 0.158 M x 0.02287 mL
= 0.003613 Moles of KOH
Step 2: Moles of H2SO4
Mole ratio; from the equation, 1 mole of H2SO4 requires 2 moles of KOH
Therefore;
0.003613 moles of KOH requires
= 0.003613/2
= 0.001807 Moles
Step 3: Molarity of H2SO4
Molarity = Number of moles / Volume
Number of moles = 0.001807 moles
Volume = 20 mL or 0.020 L
Therefore;
Concentration = 0.001807/ 0.020
= 0.0904 M
Hence, the concentration of the acid, H2SO4 was 0.0904 M or 0.0904 moles/L
Keywords: Molarity, Moles, Volume
Learn more about:
- Molarity: https://brainly.com/question/9324116
- Relationship between moles, molarity and volume: https://brainly.com/question/9324116
Level: High School
Subject: Chemistry
Chapter: Mole and stoichiometry equations
