Question

1.Determine the pH and pOH for each of the following solutions:

[OH−] = 1.6×10⁻⁷ M

[H3O+] = 4.5×10⁻⁴ M

[H3O+] = 2.0×10⁻⁴ M

[OH−] = 7.5×10⁻⁹ M

2.Acetic acid has a Ka of 1.8×10⁻⁵

What is the pH of a buffer solution containing 0.20 M HC2H3O2 (acetic acid) and 0.20 M C2H3O2−?

3. A volume of 40.0 mL of a 0.850 M HNO3 solution is titrated with 0.780 M KOH. Calculate the volume of KOH required to reach the equivalence point.

Express your answer to three significant figures, and include the appropriate units.

Answer 1

[OH−] = 1.6×10⁻⁷ M

POH =-log[OH-]

          = -log1.6*10^-7

          = 6.7958

PH   = 14-POH

        = 14-6.7958   = 7.2042

[H3O+] = 4.5×10⁻⁴ M

PH   = -log[H3O+]

       = -log4.5*10^-4

       = 3.3467

POH = 14-PH

           = 14-3.3467

          = 10.6533

[H3O+] = 2.0×10⁻⁴ M

PH = -log[H3O+]

         = -log2*10^-4

        = 3.6989

POH = 14-PH

            = 14-3.6989 = 10.3011

[OH−] = 7.5×10⁻⁹ M

POH = -log[OH-]

          = -log7.5*10^-9

          = 8.1249

PH   = 14-PH

        = 14-8.1249   = 5.8751

2.

PKa = -logKa

           = -log1.8*10^-5

           = 4.75

PH = Pka + log[C2H3O2^-]/[HC2H3O2]

       = 4.75 + log0.2/0.2

       = 4.75

3. KOH + HNO3 -----------> KNO3 + H2O

1 mole    1 mole

     KOH                                                       HNO3

M1 = 0.78M                                               M2 = 0.85M

V1   =    V2 = 40ml

M1V1    = M2V2

V1    = M2V2/M1

= 0.85*40/0.78      = 43.58ml >>>>>answer