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Topic: PH calculation  (Read 5550 times)

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Offline taylor

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PH calculation
« on: June 12, 2009, 02:49:03 AM »
hey guys ive got a question.

calculate the formate ion concentration and pH of a solution that is 0.05m in formic acid (HCOOH;Ka=1.8x10^-4) and 0.10M in HNO3.

'I'  would think when wrting dissociation expression it would go like this x/(0.05*0.1)=1.8e-4 however it goes like this 0.1x/0.05=1.8e-4. why is this?
and when i calculated the Ph i just used the concentration of the HNO3, should i include the formic acid. thanks for any help

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Re: PH calculation
« Reply #1 on: June 12, 2009, 04:27:14 AM »
x/(0.05*0.1)=1.8e-4

Please eleaborate.

Write reaction equation, write reaction quotient using real ions and molecules concentrations, explain your assumptions.
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Offline taylor

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Re: PH calculation
« Reply #2 on: June 12, 2009, 08:55:36 PM »
umm. the question says a solution is made from HNO3 and HCOOH. so i would write the eqation thing like this..

HNO3+HCOOH> H30+NO3+COO
this would mean i would get a reaction quoient of H30/(HNO3)(HCOOH) however the answer has the reaction quoient llike this (HNO3)(H30)/(HCOOH), that means the reaction equation must go something like this HCOOH>HNO3+H or sumthing?? im confused. could you show me the correct reaction eqaution please. thanks

Offline plankk

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Re: PH calculation
« Reply #3 on: June 13, 2009, 03:01:13 AM »
HNO3 is strong acid so you can assume that it dissociate fully (write the equation of nitric acid's dissociation). HCOOH is rather weak acid so write the equilibirum between the ions and molecule. (the second equation).

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Re: PH calculation
« Reply #4 on: June 13, 2009, 05:12:23 AM »
HNO3+HCOOH> H30+NO3+COO

Ignoring fact that you have ignored charges (which you should not) this is wrong - these are two separate, independent reactions. You can't mix them in one equation.
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