Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: tjpj77 on September 05, 2012, 04:40:08 AM
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Find pH if 3 mL 0.1M HCl is added to 97 mL of pure H2O at pH 7.0.
(3mL/100mL)(0.1 mol HCl/1L) = 3 x 10^-3 mol HCl/L
3 - log 3 = 2.5
pH = 2.5
right?
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Find pH if 3 mL 0.1M HCl is added to 97 mL of pure H2O at pH 7.0.
(3mL/100mL)(0.1 mol HCl/1L) = 3 x 10^-3 mol HCl/L
OK
3 - log 3 = 2.5
What are you doing here? Where does this come from?
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after you have the M of a strong acid or base, you can just do the quick math via that trick: bring down and cross over the power and then subtract from that the log of what was the first number.
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Hardly a trick, just using log properties:
log(3×10-3) = log(3) + log(10-3) = ~0.5 - 3
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Ah OK. I agree with 2.5, I was just wondering where that calculation came from. I've never bothered to expand the log before calculating for questions like this.
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For me it was obvious.
Most likely that shows the difference between me and the calculator generation :P
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I was thinking similarly ;)