Course Home 13201 SP18: MyLi × seld-14536635&Open; umWMAC-f54b45961f6881bcfSab96

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Course Home 13201 SP18: MyLi × seld-14536635&Open; umWMAC-f54b45961f6881bcfSab966330e8eda3 #10001 oncentration and Stoichiometry- Chromium nyct/itemView?assignmentProblemID-966023648toffset-next and Stoichiometry G (c) 110f 13 (> Part A Work A volume of 700 mL of aqueous potassium hydroxide (KOH) was strated against a standard soluton of sulturic acid H2SOs) What was the molarity of the KOH solution if 15.7 mI. of 1.50 M H,SO4 was needed? The equation is Policy Express your answer with the appropriate units View Available Hints) molarity alueUnits Submit PartB Radox strations are used to detemine the amounts of odizing and reducing agents in solubon. For example. a solution of bydrogen peroxide H2Op can be ttrated against a solusion of potassium permanganate KMnOs The tolloving equation represents the reaction 7:33

Explanation / Answer

Ans. Part A: Balanced reaction:    2 KOH + H2SO4 ----------> K2SO4 + 2 H2O

At equivalence point, the total number of moles of H+ from acid is equal to total number of OH- from base.

That is-           x(M1V1), acid = y(M2V2), base             - equation 1

            Where, x = moles of H+ produced per mol acid = 2 for H2SO4

                        y = moles of OH- produced per mol base = 1 for KOH

                        V and M are volume and molarity of respective solution.

# Given, Volume of acid = 15.7 mL

            Molarity of acid = 1.50 M

            Volume of KOH = 70.0 mL

            Molarity of KOH = M2

# Putting the values in equation 1-

            2 x (1.50 M x 15.7 mL) = 1 x (M2 x 70.0 mL)

            Or, M2 = (2 x 1.50 M x 15.7 mL) / 70.0 mL

            Hence, M2 = 0.673 M

Therefore, molarity of KOH solution = 0.673 M

Note: Half of data in Part B is missing (not visible). Use can use the above expression accordingly to solve part B, too.

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