Consider an acid, HA, with pKa = 5.883. 0.19 moles of hydroxide ion are added to 61.99mL of a 0.412M solution of the acid. What is the resulting pH?
Answer is 3.758. Thanks for your help.......(by the way, I have a test on Wednesday so yeah.........lol)
2006-10-23 19:36:16 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Coupla things mate
You added 0.19 millimoles (mmol) of OH
The reaction is HA + OH- ===> H2O + A-
Adding 0.19 mmol of OH- will yield 0.19 mmol of A- or 0.19/61.99 = 3.065 x 10^-3 mol/l
THe HA by difference is 0.412 - 3.065x10^-3 = .4089 mol/l
Ka = [H+] [A-] / [HA]
Rearranging and taking logs
pH = pKa + log([A-] / [HA])
= 5.883 + log(3.065x10^-3/0.4089)
= 3.758 (to 4 sig figures)
Good luck with your test. May the force be with you!
2006-10-23 21:19:47 · answer #1 · answered by Anonymous · 1⤊ 0⤋
The answer is WRONG.
In your solution you have moles HA =M*V = 0.412 *61.99*10^-3= 0.0255
It is a monoprotic acid, so it will react 1:1 with the OH-.
If you truly added 0.19 mole OH-, then the base is in excess and the pH will be alkaline.
You have 0.19-0.0255=0.1645 mole OH- remaining which corresponds to a concentration of [OH-] = 0.1645/(61.99*10^-3)=2.65 M.
It is so concentrated that you go out of the pH scale (pH=14.42) and you don't even need to bother about the weak conjugate base (A-) that is formed.
Check the numbers that you provide in the question.
By the way, the assumptions made to use the Henderson-Hasslbalch do not allow you to give the pH value with an accuracy of 3 decimal digits. Quoting the limitations as mentioned in wikipedia
"There are some significant approximations implicit in the Henderson-Hasselbalch equation. The most significant is the assumption that the concentration of the acid and its conjugate base at equilibrium will remain the same as the formal concentration. This neglects the dissociation of the acid and the hydrolysis of the base. The dissociation of water itself is neglected as well. These approximations will fail when dealing with relatively strong acids or bases (pKa more than a couple units away from 7), dilute solutions (1 mM or less), or heavily skewed acid/base ratios (more than 100 to 1)."
Normally pH values are given with 2 decimal digits and if you want to have an exact solution, you need to solve the equilrium problem expilictly.
2006-10-24 05:34:16 · answer #2 · answered by bellerophon 6 · 0⤊ 1⤋
pH = pKa + log(conc of salt/conc of base)
since the acid is limiting reagent,,,, the conc of salt is equal to the number of moles of acid/ volume.
conc of base = number of moles given- reacted/volume.....
2006-10-24 02:45:04 · answer #3 · answered by !kumar! 2 · 0⤊ 1⤋