Silver sulfate solution is added to 25.00 mL of a 0.500 M potassium chloride sol

Question

Silver sulfate solution is added to 25.00 mL of a 0.500 M potassium chloride solution until no more precipitate forms. What mass of silver chloride will be formed? What concentration of potassium will remain in solution after the reaction is complete if 45 mL of silver sulfate solution were added to the potassium chloride solution? Silver sulfate solution is added to 25.00 mL of a 0.500 M potassium chloride solution until no more precipitate forms. What mass of silver chloride will be formed? What concentration of potassium will remain in solution after the reaction is complete if 45 mL of silver sulfate solution were added to the potassium chloride solution?

Explanation / Answer

2KCl(aq) + Ag2SO4(aq)----> 2AgCl(s) + 2KNO3(aq)
2 mole                   2 mole

no of moles of KCl = Molarity * volume in ml
                    = 0.5 * 0.025
                    = 0.0125 moles of KCl
from balanced equation
2 mole of KCl react with Ag2So4 to from 2 mole of AgCl
0.0125 mole of KCl react with AgNO3 to from 0.0125 moles of AgCl
mass of AgCl = no of moles of AgCl * molar mass of AgCl
             = 0.0125*143.4 = 1.793 gm of AgCl

KCl
molarity M1   = 0.5 M
volume V1     = 25ml
no of mole n1 = 2

Ag2SO4
molarity M2    =
volume    V2   = 45ml
no of moles n2 = 1

M1V1/n1    =   M2V2/n2

M2          = M1V1n2/n1V2
            = 0.5*25*1/2*45
            = 0.138M
concentration of KCl left = 0.5-0.138 = 0.368 M

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