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pH


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#1 cobainfan

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Posted 03 May 2008 - 06:44 PM

Hi,

I'm having troulbe calculating the pH of 5*10^(-2) moles per litre Sodium Hydroxide. Any help would be much appreciated.

#2 Marcus

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Posted 04 May 2008 - 11:21 AM

Is this in higher??? blink.gif blink.gif

right, in water: [H+][OH-]=10-14 ([] means concentration of)

rearranging this you get [H+]= \frac{10^{-14}}{5\times 10^{-2}}
so [H+] = 2 \times 10^{-13}

now I don't know of any higher method of working out the pH from here. so this is adv. higher

pH=-log_{10}{[H^{+}]}
so pH would be 12.7


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#3 Linzi2011

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Posted 08 May 2008 - 02:08 PM

QUOTE(Marcus @ May 4 2008, 12:21 PM) View Post
Is this in higher??? blink.gif blink.gif

right, in water: [H+][OH-]=10-14 ([] means concentration of)

rearranging this you get [H+]= \frac{10^{-14}}{5\times 10^{-2}}
so [H+] = 2 \times 10^{-13}

now I don't know of any higher method of working out the pH from here. so this is adv. higher

pH=-log_{10}{[H^{+}]}
so pH would be 12.7


Aparrently so, we had a few questions on this in class about calculating the pH of a solution.

Its simple really once you know how it works as Marcus has said laugh.gif

#4 Xeraii

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Posted 11 May 2008 - 06:29 PM

This definitely isn't in the Higher course. Your teacher must just be giving you some extra problem solving questions. smile.gif





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