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Answer #1

             HCl(aq) + KOH(aq) ----------------> KCl(aq) + H2O(l)

           1 mole      1 mole

     HCl                                                            KOH

M1   =                                                           M2 = 4.96*10^-2M

V1   = 11.9ml                                                 V2 = 35.3ml

n1   = 1                                                          n2 = 1

                M1V1/n1    =      M2V2/n2

                   M1            =    M2V2n1/V1n2

                                     =   4.96*10^-2 *35.3*1/11.9*1    = 0.147M

The molarity of HCl   = 0.147M

part-B

2HClO4(aq) + Ba(OH)2(aq) -----------------> Ba(ClO4)2(aq)   + 2H2O(l)

2 mole                1mole

         HClO4                                                 Ba(OH)2

         M1= 0.326M                                           M2 =

           V1   = 15.7ml                                         V2 = 30.1ml

             n1 = 2                                                 n2 = 1

                             M1V1/n1    =    M2V2/n2

                                  M2        = M1V1n2/V2n1

                                                  = 0.326*15.7*1/(30.1*2)

                                                 = 0.085 M

The molarity of Ba(OH)2   = 0.085M

part-C

PH    = PKa + log[CN^-]/[HCN]

9.12    = 9.4 + log[CN^-]/[HCN]

log[CN^-]/[HCN] = 9.12-9.4

log[CN^-]/[HCN]    = -0.28  

[CN^-]/[HCN]         = 0.525/1

part-D

PH    = PKa + log[HCO3^-]/[H2CO3]

6.18    = 6.38 + log[HCO3^-]/[H2CO3]

log[HCO3^-]/[H2CO3] = 6.18-6.38

[HCO3^-]/[H2CO3]    = 0.63/1

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