A solution of HCl is prepared by diluting 25.0 mL of a 1.0 M HCl solution...

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A solution of HCl is prepared by diluting 25.0 mL of a 1.0 M HClsolution with enough water to make 750 mL of HCl solution. (Showyour work for all calculations!)

a) What is the molarity of the HCl solution?

b) What is the [H3O+] and the pH of the HCl solution?

c) Write the balanced chemical equation for the reaction of HCland Ba(OH)2

d) How many milliliters of the diluted HCl solution is requiredto completely react with 350 mg of Ba(OH)2?

Answer & Explanation Solved by verified expert
3.8 Ratings (641 Votes)

before dilution                                       after dilution

M1 = 1M                                                  M2

V1 = 25ml                                                 V2 = 750ml

            M1V1    = M2V2

             M2    = M1V1/V2

                      = 1*25/750 = 0.034M

HCl + H2O---------> H3O+ + Cl-

0.034 M       0.034M

[H3O+] = [HCl]

[H3O+]   = 0.034M

PH = -log[H+]

         = -log0.034   = 1.4685

Ba(OH)2 + 2HCl --------> BaCl2 + H2O

no of moles of Ba(OH)2 = W/G.M.Wt = 0.35/171.34 = 0.00204moles

Ba(OH)2 + 2HCl --------> BaCl2 + H2O

1 moles of ba(OH)2 react with 2 moles of HCl

0.00204 moles of Ba(OH)2 react with = 2*0.00204/1   = 0.00408moles of HCL

molarity = no of moles/volume in L

0.034      = 0.00408 /volume in L

volume in L = 0.00408/0.034 = 0.12L = 120ml >>> answer


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