percent protonation when the percentage is greater than 5
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percent protonation when the percentage is greater than 5
can someone help explain what is means when the percent protonation is higher than 5%? why does that mean that we have to go back and solve using the quadratic formula even though the Ka value is less than 10^-3?
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Re: percent protonation when the percentage is greater than 5
I think it's just a way of checking the validity of a calculation. If our x-value is more than 5% of the original, we have to make sure that we go back and re-do the calculation.
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Re: percent protonation when the percentage is greater than 5
If the x value that we solve for is greater than 5% of the initial concentration, then the difference that we assumed was negligible ( say 0.25 - x ) is actually not. Thus, we have to use the quadratic formula to find the value of x. The approximation assumes the x is so small that the equilibrium concentrations of the initial concentrations we were given essentially are the same as the initial ones. Thus, x > 5% of the initial concentration would make this assumption false. Also, it's usually safe to use the approximation for when K is less than 10^-3 ( so 10^-4 or less to be safe), but it doesn't hurt to check.
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Re: percent protonation when the percentage is greater than 5
The reason we do the 5% check for percent protonation is simply to make sure that doing small x approximation was the correct choice. It is used as a like a second line of defense to make sure it is okay to consider x as negligible. If you get x as greater than 5% you should go back and resolve the equation for equation using the quadratic formula since x would no longer be negligible. If lower than 5%, your decision to approximate was correct.
Re: percent protonation when the percentage is greater than 5
This is just a way to check whether making assumptions when solving the equation is ok. When the percent protonation is higher than 5% it means that the x value does have some significance. This means we can't make assumptions and that we have to go back and do the quadratic equation.
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