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Calculating PH using pk?
The pk of an acid is 4.16: Calculate the ph of a solution titrated with 63% NaOH .... I was doing the calculations and I got 4.39 but I was not sure if I just subtracted 63% or .63 from 100 to get the conjugate acid concentration [HA]? Can anyone help?
3 Answers
- 9 years agoFavorite Answer
you need this tutorial
- Anonymous4 years ago
I fairly have now now no longer tried your first query by way of certainty that i do now no longer see how this might make an acid buffer. Please look at which you have have been given have been given submitted the query effectively. the only acid buffer determination is D a million.23 and this mixture can no longer furnish the kind of pH 2d query: The pH of a buffer is given indoors the process the Henderson - Hasselbalch equation: pKa HCOOH = - log Ka = -log( a million.seventy seven*10^-4) pKa = 3.seventy 5 pH = pKa + log ([salt]/[acid] pH = 3.seventy 5 + log (0.89/a million.12) pH = 3.seventy 5 + log 0.795 pH = 3.seventy 5 + ( - 0.10) pH = 3.seventy 5 -0.10 pH = 3.sixty 5 determination B
- ?Lv 79 years ago
pKa = 4.16 so Ka = 10^-pKa = 6.92x10^-5
Ka = [H+][A-] / [HA]
63% OH- = 63g of 100g NaOH solution = OH-
63g / 40g/mole = 1.58moles NaOH
assuming equivalence point with NaOH titration, moles OH- used = moles H+ neutralized
1.58moles OH- used neutralized 1.58moles H+ from acid
what are the volumes? we need volumes for concentrations, not just moles
