% Protonation

[Natalie Swerdlow 3K]

Why do we have to use the quadratic equation for calculating % protonation if the Ka value is less than 10^-3 and x is negligible for calculating the concentrations? Specifically for Achieve question 3 I had to use the quadratic equation only for % protonation but calculated the value of H+ for pH assuming x was negligible.

[Raymond_Kallely_1A]

I believe that the idea is that x is so small that it doesn't change the equilibrium constant(Ka). On the other hand, the way I see it is that percent ionization acts as a ratio of sorts between the two sides of the equation, meaning that the quadratic equation is needed to solve for it.

[Ayaan_Ekram_2J]

Hello!

Even when you assume x is negligible due to the small value of K, you must then check if the value of x is greater than 5% of the the initial value of the given substance. If the x is greater than 5%, you cannot assume x is negligible and must again use the quadratic equation as x will have an effect on the value of the equilibrium concentration of the reactant. I hope this helps!

[Sujin Lim 1H]

When x is very small, we say that it s negligible, but ONLY in respect to the initial reactant concentrations. This does not mean that there is no change in concentrations during the chemical reaction, because some amount of product must be formed from the reactants if the reaction needs to reach equilibrium. Therefore, we can approximate the equilibrium reactant concentrations as nonchanging from their initial concentrations, but we need to still figure out how much product was formed, no matter how small that amount may be.

[905767718]

The only time that it's ok to estimate that x is negligible is when the percent protonation is less than 5% because that means that the amount was small enough that saying x was zero would have almost no effect on the total change of the initial concentration of the reactant.