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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.

## Want to join the conversation?

• 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?
• Because you're multiplying the x outside the bracket (see ), not adding or subtracting. Multiplying or dividing by 0.35 is considerable compared to adding or subtracting a tiny value from 0.35.
• at , what is the dissociation constant of CH3COONa? How do we know that it completely dissolves and there is 1M CH3COO- and 1M Na+?
• When an ionic bond breaks down, it dissociates completely. When a compound breaks down in an acid/base reaction, it may not be complete because the water is needed to pull off the H+ to form hydronium, but in the case of Na+ it will completely fall off
• ***
In the first example, why doesn't the Na+ get into the equilibrium formula?
Like:
CH3COOH + H2O <---> (CH3COO-) + (H3O+) + (Na+)
• The sodium is known as a "spectator ion." This means that it does not react with anything but remains present in the solution. The correct way to represent the equation would be:
CH3COOH + H2O + (Na+) <---> (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.
• At why is the [NH3} a reactant instead of a product. Is it because we are using Kb instead of Ka?
• 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?
• why doesn't NO3 in NH4NO3 react with anything?
• NO3- is the conjugate base of a strong acid (HNO3), that means it’s very unfavourable for it to react with water.
• 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-?
• As long as the amount of correct OH- or H3O+ is calculated, then the result will always be the same.
• 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?
• Sodium acetate is completely soluble in water, right? So if we add acetate to water, the following equation happens: { CH3COOH + H2O <==> CH3COO- + H3O+ }. You know this already, because you said the equilibrium shifts to the left, which is correct. So, if I'm trying to take CH3COO- and turn it into CH3COOH, I need to pull one of the hydrogens off of hydronium, and stick it on the CH3COO- to make CH3COOH. Then I end up with CH3COOH and H2O. This decreases the concentration of hydronium, because we turned a bunch of hydronium into water. Hopefully that helps!