Calculate the volume of .100M NaOH that must be added to reach ph of 3 in the titration of 25.00 mL of .100M HCl

Answer 1

mols HCl initially = 0.025 x 0.1 = 0.0025
To end up with pH = 3 (0.001M H^+) we want H^+ to be 1E-3. If we work in millimols and let x = mL of 0.1M NaOH, we have then
[(25.00 x 0.1M)-(0.1x)/(25+x)] = 1E-3M
Solve for x and I obtained approx 24.5 mL of 0.1M NaOH that must be addd. We can check to see if that is right and we have
25.00 x 0.1M HCl = 2.5 mmols initially.
-24.50 x 0.1M NaOH = 2.45mmols NaOH add
-------
0.5 millimols HCl reamins. (HCl) = 0.5 mmol/(24+24.5 mL) = 0.00101
for which pH = 2.996. If you carry the numbers another place or two in the calculations above you will get 1E-3 and pH 3.