Recent content by Comrade

A-ha! Okay, here it is. after addition of 10mL final volume is .035L. So 6.3*10^-4 / .035L = .018 M [A-] .00116 moles HA - 6.3*10^-4 = 5.3*10^-4 moles (HA left in flask). 5.3*10^-4 / .035 = .015 M [HA] then: solving for pKA, pKa= 5.87 - log(.018/.015) pKa= 5.79 Ka = 10^-5.79 Ka =...

Alright, I just keep repeating my same thought pattern or something, because I keep coming up with: .0633 M x .01L = .000633moles NaOH Someone save me from my sinful ways! edit: woah wait a minute. okay, so why am i using the 18.4 mL of base still? if it's asking for Ka after the addition of...

I calculated for moles in 10mL of NaOH to be 6.3*10-4 1 mol of NaOH = 1 mol of monoprotic acid, moles monoprotic acid = 6.3*10-4 then, using: pH = pKa + log\frac{[A-]}{[HA]} If my reasoning is correct, log\frac{[A-]}{[HA]} will equal 1 since [A-] and [HA] will equal the moles of HA...

I've been trying to figure this one out for a while. A 0.1276g sample of a monoprotic acid was dissolved in 25mL H2O, and titrated with a .0633 M NaOH solution. Volume required to reach equivalence point was 18.4 mL. It asked me to calculate the molar mass, which i found to be 116g/mol: 0.0633...