Question

3. What is the pH of the resulting solution after 150 mL of a 3.2 M sodium hydroxide solution is mixed with 4.0 mL of a 4.0 M vinegar solution?

4. In the above question is the acid or base a limiting reagent?

5. Calculate the pH after adding 40.0 mL of 3.7 M potassium hydroxide to 20.0 mL of 12 M formic acid.

6. How many more mL of the 3.7 M potassium hydroxide solution must be added to the formic acid solution to reach the equivalence point in the above question (#5) above?

7. Calculate the pH of a solution made by adding 70 mL of 4.3 M sodium acetate to 30 mL of 10 M acetic acid.

*Help with any of these could be so wonderful; please show/tell steps*

Answer 1

3)

no of mol of NaoH added = 150*3.2/1000 = 0.48 mol

no of mol of Vinegar(aceticacid) = 4*4/1000 = 0.016 mol

limiitng reactant = vinegar(aceticacid)

pH = pka + log(base/(acid-base))

   pka of aceticacid = 4.74

   pH = 4.74 + log(0.016/(0.48-0.016))

      = 3.28

4) limiitng reactant = vinegar(aceticacid)

5) no of mol of KoH added = 40*3.7/1000 = 0.148 mol

   NO of mol of formicacid = 20*12/1000 = 0.24 mol

pH = pka + log(base/acid)

pka of formic acid = 3.75

pH = 3.75+log(0.148/(0.24-0.148))

     = 3.96

6) NO of mol of formicacid = 20*12/1000 = 0.24 mol

     NO of mol of KOH required to reach equivalence point = 0.24 mol

      volume of KOH required = n/M = 0.24/3.7 = 0.065 L

                              = 65 ml

    Volume of extra KOH required = 65 - 40 = 25 ml

7) no of mol of sodium acetate = 70*4.3/1000 = 0.301 mol

   no of mol of aceticacid = 30*10/1000 = 0.3 mol

equivalence point

   pH = pka +log(sodium acetate/aceticacid)

   pH = pka +log(0.3/0.3)

   pH = pka = 4.74