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umenie Will Due to run if you leave the test. Remaining Time: 39 minutes, 49 seconds. Less than half of the time rer Question Completion Status: Close Window Moving to another question will save this response. Question 17 of 25 Save and Question 17 4 points In a titration experiment, 974 mL of 0.696 M KOH was needed to neutralize 19 mL H2SO4 solution. What is the concentration (M) of the original sulfuric acid solution (Answer in two decimal places) 2KOH + H2SO4 --Na2SO4+ 2H20 Moving to another question will save this response. Question 17 of 25 Close Window

Answer 1

* Solution Given Date: Koh - Vi= 9thmi Mi = 0.696 M Hason 7 ₂ = 19 mi - M2 = 2 Reaction : 2KOH + Hason » Nasoy + 2H₂O

moles can be written as : moles - molarity (M) X volume (1) ( This is because m = mules/ moles = MY component a & As 4 we have conside kol as H₂son as component az M . a moles moles of of Kolt H₂son' = ,V, M2 V2

=> from the reaction Stoichiometry www wewny for a mole koh = Required a mole of Hason for Miv, mole of koll & Mirixi moles of Hason . moles of H₂son = Mini XI We also know: moles of Hason = M₂V, - 2) L. from 040 , v, x) - MV - Mr = Mivo - 0.696(M) X 974 (mi) - 2x 19 (mi) M₂ = 17.83 M l ) of concentration sulfuric acid.