Chemistry, 16.09.2019 13:00 ayoismeisjjjjuan

The titration of a 20.0 ml sample of an h2so4 solution of unknown concentration requires 22.87 ml of a 0.158 m koh solution to reach the equivalence point. what is the concentration of the unknown h2so4 solution?

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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: Relationship between moles, molarity and volume:

Level: High School

Subject: Chemistry

Chapter: Mole and stoichiometry equations

Answer from: Quest

answer is a

explanation:

Answer from: Quest

10.063g / 10.00 ml = 1.0063 g/ml which rounds off to 4 sig figs as: 1.006 g/ml

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