You are titrating 35.00 mL of a 1.20 M solution of HCOOH with a strong base. If you add 1.00 mL of a 3.60 M solution of sodium hydroxide, what is the final pH of the acid solution? The Ka for HCOOH is 1.8 X 10-4.

Answer 1

begin with 35.00 mL x 1.20 = 42 mmol.
add 1.00 mL x 3.60 M NaOH = 3.60 mmol.
..........HCOOH + NaOH ==> HCOONa + H2O
begin.....42 mmol...0........0.......0
change....-3.60.....3.60....3.60.....3.60
final......38.40....0.......3.60.....3.60

You see the final solutin is a buffer; i.e., it has a weak acid (HCOOH) and a salt of the weak acid(HCOONa) present.
Use the Henderson-Hasselbalch equation.
pH = pKa + log[(base)/(acid)]
Use pKa of HCOOH
(base) = (HCOONa)
(acid) = (HCOOH).
Plug and chug. Post your work if you get stuck.

Answer 2

If a sample of oxygen-15 has an activity of 8000 becquerels, how many minutes will elapse before it reaches an activity of 500 becquerels?