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# Common ion effect and buffers

The common ion effect describes the effect on ​equilibrium that occurs when a common ion (an ion that is already contained in the solution) is added to a solution. The common ion effect generally decreases ​solubility of a solute. It also can have an effect on buffering solutions, as adding more conjugate ions may shift the pH of the solution.

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• At , why is the x in the numerator (outside the bracket) left while the other x's are neglected taking into consideration that all the values of the x are very small and are neglected everywhere else? • at , what is the dissociation constant of CH3COONa? How do we know that it completely dissolves and there is 1M CH3COO- and 1M Na+? • ***
In the first example, why doesn't the Na+ get into the equilibrium formula?
Like:
CH3COOH + H2O <---> (CH3COO-) + (H3O+) + (Na+) • At 3.56, he accounted for the 1M CH3COOH by subtracting x, but didn't account for the 1M for CH3COO-. Over time, wouldn't the reverse reaction be more favorable and require it be included in the ICE equation? • After some time, when the reaction approaches equilibrium, the reverse reaction will start having a larger effect until it is equal to the forward reaction, but it will never go beyond this equilibrium (if you don't add any additional chemicals.)
When you continue watching you can see what he does with the formula. I_, the initial concentrations, are under the fraction, while _E, the final concentrations, are written above the fraction. The reason why H2O is excluded is because water doesn't have concentration (it's "dissolved" in itself in some way)
• At why is x assumed to be much smaller than 1? How would I know to make this assumption in other problems? • Chemists are lazy creatures.
They assume that x≪1 in order to avoid having to solve a quadratic equation.
But they always have to check whether the assumption is valid.
A common rule of thumb is that x is negligible if the initial concentration of HA divided by Ka is greater than 400.
• Why isn't the reaction between water and NO3- taken into account? Doesn't NH4NOE dissociate, and cause NO3- to raise the concentration of OH- since it acts as a base, and thus change the initial concentration of OH- on the ICE table to 0.35M? • For the second example problem pertaining NH3 and NH4+NO3-, instead of having the NH3 react with water to form NH4+ and -OH, I had NH4+ react with water to form H3O+ and NH3.

NH4+ + H2O <-----> H3O+ + NH3
Initial concentration = 0.35 nothing 0 0.15
Change in concentration = -x nothing +x +x
End concentration = 0.35-x nothing x 0.15+x

I then calculated Ka using (Ka)(Kb)=(Kw), with the given Kb value. Ka = 5.56x10^-10 = [h3o+][nh3]/[nh4+] = (x)(.15+x)/(.35-x).... and solve for x, which = 1.3x10^-9; then solve for pH using x = 8.88; which is exactly what was obtained in the video.

Question is, will you also obtain the correct pH value regardless which compound you use to react with water as long as you set up the reaction equation properly to either form H3O+ or -OH and use the Ka value for H3O+ and Kb value for -OH in order to solve for x and then further solve for pH or pOH? Or do you have to use the molecule that doesn't form a by-product; CH3COOH vs CH3COONa, Na+ being considered the by-product, therefore CH3COOH favored; likewise, NH3 favored over NH4+ with by-product NO3-? • How does adding sodium acetate increase the concentration of acetate ion and decrease the hydronium concentration. I understand the equilibrium shifted to the left because of le chatliers rule but what does that have to do with the change in concentrations?   