t Titrations and Solution Stoichiometry A titration is a procedure for determini

Question

t Titrations and Solution Stoichiometry A titration is a procedure for determining the concentration of a solution by allowing it to react with another solution of known concentration called a standard solution). Acid-base reactions and oxidation-reduction reactions are used in titrations. For example, to find the concentration of an HCl solution (an acid), a standard solution of NaOH (a base) is added to a measured volume of HCl from a calibrated tube called a buret An indicator is also present and it w change color hen all the acid has reacted. Using the concentration of the standard solution and the volume dispensed, we can calculate molarity of the HCl solution. Part A A volume of 60.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). WI was the molarity of the KOH hat solution if 14.7 mL of 1.50 MH2SO4 was needed? The equation is 2KOH(aq) H2SO4 (aq)- K2So, (aq) 2H2O(l) molarity Value Units Part B Redox titrations are used to determine the amounts of oxidizing and educing agents in solution. For example, a solution of hydrogen H2O2, can be titrated against a solution potassium permanganate, The following represents the reaction O2 (g) +2MnSO4 (aq)+ K2SO4 (aq) 4H2O(l) A certain amount of hydrogen peroxide was dissolved in 100. mL of water and then titrated with 1.68 MKMno. hat mass of HoO was dissolved if the titration required 16.3 mL of the KMno4 solution? mass of H202 Value Units

Explanation / Answer

KOH H2SO4

M1 = ? M2 =1.5 M

V1 = 0.060 L V2 = 0.0147 L

n1 = 2 n2 = 1

M1V1 = M2V2

n1 n2

M1 = M2V2 * n1

V1 * n2

M1 = 0.735 M [answer to part A]

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part B

H2O2 KMnO4

M1 = ? M2 = 1.68 M

V1 = 0.1 L V2 = 0.0163 L

n1 = 1 n2 = 2

M1 = M2V2 * n1

V1 * n2

= 0.137 M is the concentration of H2O2

That means 0.137 moles are present in 1 litre

Moles = amount in grams of H2O2

Molar mass of H2O2

Amount = Moles * Molar mass

= 0.137 * (34 g/mole)

= 4.66 g [answer to part B]

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