Question

A) A 25 mL sample of 0.723M HClO4 is titrated with a 0.273M KOH solution. What is the pH before any base is added?

B) A 25 mL sample of 0.22M hydrazoic acid (HN3 ; Ka=2.6e-5) is titrated with a 0.30M KOH solution. What is the pH of the solution after 16 mL of base is added?

C) A 25 mL sample of an HCl solution is titrated with a 0.15M NaOH solution. The equivalence point is reached with 75 mL of base. The concentration of HCl is _____M.

Answer 1

A. HClO4 ------------> H^+ (aq) + ClO4^-

0.723M                     0.723M

[H^+]   = [HClO4]

[H^+]   = 0.723M

PH = -log[H^+]

      = -log0.723 = 0.14086

B. no of moles of N3H   = molarity * volume in L

                                     = 0.22*0.025   = 0.0055 moles

no of moles of KOH    = molarity * volume in L

                                    = 0.3*0.016   = 0.0048moles

                    N3H    +     KOH --------------> N3K + H2O

I                   0.0055       0.0048                   0

C                 -0.0048    -0.0048                   0.0048

E                 0.0007        0                        0.0048

           Ka = 2.6*10^-5

           PKa = -logKa

          PKa = -log2.6*10^-5

                    = 4.5850

      PH   = PKa + log[N3K]/[N3H]

              = 4.5850 + log0.0048/0.0007

             = 4.5850+0.8316   = 5.4166

c.    HCl + NaOH ---------------------------> NaCl + H2O

     1 mole   1 mole

HCl                                                                    NaOH

M1 =                                                                  M2 = 0.15M

V1 = 25ml                                                         V2   = 75ml

n1 = 1                                                                n2 = 1

                M1V1/n1       =      M2V2/n2

                  M1                =   M2V2n1/V1n2

                                      =   0.15*75*1/25*1   = 0.45M

conc of HCl     = 0.45M