I am adding 0.55 mL of 0.1M NaOH to 10 mL of 0.5M Formic Acid.
Here are the values I calculated to use.
Moles H+ : 0.5 * (10mL / 1000) = 0.005 mol
Moles OH- : 0.1 * (.55mL / 1000) = 0.000055 mol
volume : (10mL / 1000) + (0.55mL / 1000) = 0.01055 L
Moles Total : moles H - moles OH = 0.005 - 0.000055 = 0.004945 mol
M Total : moles Total / volume = 0.004945 mol / 0.01055 L = .46872 M
Moles Salt : 0.000055 mol
M Salt : moles salt / volume = 0.000055 mol / 0.01055 L = 0.00521 M
Calculations to find the pH
pH = -log(1.80E-4) = 3.745
3.745 - ( -log(M Salt / M Total) ) = 3.745 - 1.95407 = 1.791
However, this value is too low as the lowest pH of 0.5M Formic Acid is ~2.03.
If the volume of NaOH were to say, 0.5 mL, then the M Salt / M Total value would be less than 1/90, and the pH could then be found by
-log( sqrt( 1.80E-4 * M Total ) ) = 2.036
...which makes more sense, but doesn't apply to the initial problem.
Any idea of where I'm going wrong?