Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Edher on March 09, 2005, 06:22:27 PM
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Saludos,
I'm having some difficulties understanding exactly what they're asking me for.
For the titration of 50.00mL of 0.00812M Ba(OH)2 with 0.0250 M HCl, calculate
(a) the initial pH
(b) the pH when neutralization is 50% complete
(c) the pH when neutralization is 100% complete
Now, for the inital pH, I simply found the negative log of the molarity of the base
-(log 0.00812) which is 2.09. Then I subtracted this from 14.00 (since this is a base) which yielded pH 11.9. However, according to my book, the pH is 12.21.
Second, on (b) do they mean titration rather than neutralization. Because if it was neutralization, (knowing that complete neutralization yields pH 7.00) wouldn't it be possible to simply take the starting pH, (in this case my 'mistaken' one which is 11.9) Take 11.9 subtract 7.00 from it (since pH 7.00 is the complete neutralization and we want to find the pH at 50% of the neutralizationg) and then divide the result by 2 since that would give you 50% of the neutralization?
11.9 - 7.00 = 4.9 4.9/2 = pH 2.45
Thank You,
Edher