Acid HA has a Ka of 2.00 x 10-4mol dm-3. A solution was made by adding 15cm3 of 0.34 M NaOH to 25cm3 of 0.45M HA. Calculate the moles and the concentration of A- and HA in this solution. Using the expression for Ka calculate the pH of the solution
Write the equation: NaOH + HA --> NaA + H2O. Find the moles of A-: the moles of A-= the moles of NaOH added. 15 x 10-3 x 0.34= 5.10 x 10-3. Workout the initial moles of HA: 25 x 10-3 x 0.45= 0.01125 (the values are given in the question above). Workout the moles of HA remaining (as it has reacted): 0.01125-0.00510= 6.15 x 10-3. Workout the concentration of A-: there are 0.00510 moles in the solution (15+25=40cm3). To work out the concentration 5.10 x 10-3 x 1000/40= 0.1275 M. Workout the concentration of HA: using the 6.15 x 10-3 mole that we were remaining, the calculation is as follows 6.15 x 10-3 x 1000/40= 0.1538 MKa= [H+] [A-]/[HA]=2.00 x 10-4 mol dm-3. We have the following information: Ka= 2.00 x 10-4 mol dm-3, HA=0.1538M and A-=0.1275. We need to rearrange the above equation to Ka [HA]/[A-] and substitute in the numbers so we have 2.0 x 10-4 x 0.1538/0.1275=2.41 x 10-4. Workout the pH: -log (2.41 x 10-4) =3.62
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