1) What is the concentration of HCl in the final solution when diluted with pure

Question

1) What is the concentration of HCl in the final solution when diluted with pure water to a total volume of 0.15 1? 2) How many milliliters of a 9.0 M H2504 solution are needed to make solution? 0. 45 L of a 3.5 M 3) 6.74 g of the monoprotic acid KHP (MW 204.2 g/mol) is dissolved into water. The sample is titrated with a 0.703 M solution of calcium hydroxide to the equivalence point. What volume of base was used? (Hint: If a little stuck, or needing a start, go to problem 4 first! It breaks it all down!)

Explanation / Answer

1. before dilution                             after dilution
    M1 = 9M                                       M2 =
    V1 = 65ml                                     V2 = 0.15L = 150ml
             M1V1   = M2V2
               M2   = M1V1/V2
                    = 9*65/150   = 3.9M
The conc of HCl = 3.9M
2.
    before dilution                           after dilution
    M1 = 9M                                    M2 = 3.5M
    V1 =                                      V2 = 0.45L = 450ml
             M1V1    =    M2V2
                V1   =    M2V2/M1
                     =    3.5*450/9   = 175ml >>>answer

3. 2KHP + Ca(OH)2 -----------> Ca(KP)2 + 2H2O
   2 mole 1 mole
   no of moles of KHP = W/G.M.Wt
                       = 6.74/204.2 = 0.033 moles
   2 moles of KHP react with 1 moles of Ca(OH)2
   0.033 moles of KHP react with = 1*0.033/2 = 0.0165 moles of Ca(OH)2
   no of moles of Ca(OH)2 = molarity * volume in L
         0.0165            = 0.703* volume in L
      volume in L           = 0.0165/0.703 = 0.0235 L = 23.5ml >>>>answer

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