Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: merced on April 25, 2006, 11:48:12 PM
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Find the pH of a 0.000100 M solution of HOBr.
Given: Ka = 2.0 * 10-9
Using a RICE table, Ka = [OBr-][H3O+] / [HOBr]
Let [OBr-] = [H3O+] = x
I got the 2nd degree polynomial: 0 = x2 + 2.0 * 10-9 x - 2.0 * 10-13
Solving for x, I get: x = 4.462 * 10-7 = [H3O+]
Then I find that the pH = 6.350.
The answer supposedly has a pH of 6.262. What did I do wrong?
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Try to include an auto-dissociation of water to your calculations. This leads to a cubic equation.
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Then I find that the pH = 6.350.
The answer supposedly has a pH of 6.262. What did I do wrong?
Your result is correct. BATE gives pH=6.34 and calculation result includes all possible equilibria present in the solution.
Check it by yourself - link to BATE is in my signature, you will have to enter pKa (8.69) manually, as hypobromous acid is not present in the default (demo version) database.
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thanks!
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can i add to this question list?
50ml of HCl 0.0100M added to 50ml 0.0100M NaOH, what is the pH?
thnx
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Easy. Calculate how much NaCl forms (NaCl is a neutral salt since it is a salt of a strong acid and a strong base). Then you can see if there's an excess NaOH or HCl. Then you can simply calculate the pH with the formula pH= - lg(c0) since NaOH and HCl are both strong a strong base and a stong acid.
But think a bit before caculating and wasting your time...the same volume of a monoprotic base with the same concentration of the acid is added to a monoprotic acid...what could happen ;) ?
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http://www.chembuddy.com/?left=pH-calculation-questions&right=pH-of-mixture-q1