part A. a volume of 90.0mL of aqueous potassium hydroxide (KOH) was titrated aga

Question

part A. a volume of 90.0mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution od sulfuric acid (H2SO4). What was the molarity of the (KOH) solution if 25.7mL of 1.50 M H2SO4 was needed? the equation is 2KOH (aq)+H2SO4 (aq)-->K2SO4 (aq) +2H2O (l)

Part B

Resources d Solution Stoichiometry x previous | 6 of 13 ner Part A A volume of 90.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of rox sulfuric acid (H,SO4). What was the molanity of the KOH solution if 25.7 mL of 150 M H2S04 was needed? The equation is 2KOH(aq)(ag)-K,SO,(ag)+2H2o0) Express your answer with the appropriate units. molartyFalue molarity Hints wer Part B

Explanation / Answer

Part-A

We can use the equation V1*2S1 = V2*S2 Because the equation says 2 moles of KOH is required to neutralize 1 mol of H2SO4 as 2H+ will be realeased by it.

where, v denotes volume and s denotes molarity. Suffix 1 stands for H2SO4 and 2 stands for KOH

In this case, V1=25.7 mL and S1 = 1.50 M of H2SO4

and V2 = 90 mL of KOH and S2=?

S2 = 25.7*2*1.50/90 M = 0.857 M is the molarity of KOH.

Part B

Moles of KMnO4 = 1.68*21.3/1000 moles = 0.0358 moles

2 moles of KMnO4 react with 1 mole of H2O2.

So, number of moles of H2O2 reacted = 0.0358/2 moles.

Molar mass of H2O2 = 34 g/mol

So, H2O2 used was = 34*0.0358/2 = 0.609 g.

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