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### Course: Chemistry library>Unit 11

Lesson 2: Acid-base equilibria

# pH of salt solutions

The pH of a salt solution is determined by the relative strength of its ​conjugated acid-base pair. Salts can be acidic, neutral, or basic. Salts that form from a strong acid and a weak base are acid salts, like ammonium chloride (NH4Cl). Salts that form from a weak acid and a strong base are basic salts, like sodium bicarbonate (NaHCO3).

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• We consider X << 0.25 or what ever the value given in a question (assumptions). Due to this we take x as 0. Then why don't we take x square as zero?
• When we have 0.25 - x, we may assume that x is negligible in comparison to the 0.25. We are not saying that x = 0.
If x = 0.001, then 0.25 - x = 0.25 (within allowable significant figures). This is equivalent to saying that x might as well be zero.
But x² by itself still has a value. If x = 0.001, then x² = 0.000 001. This number is not being compared with 0.25. If anything, it is being compared with zero. No matter how small it is, it is infinitely larger than zero, so we cannot ignore it.
• I thought the acetate was a strong conjugate base( because acetic acid is a weak acid), I used the to the strong base way of calculate the pH. Why did Jay use the weak base formula?
• I think the 'strong base if weak conjugate acid' argument only really works if the conjugate acid is less acidic than water. In case of acetate ion, its conjugate acid CH3COOH, while definitely a relatively 'weak' acid', isn't weaker than water, so its conjugate base is a weak base.

You can calculate the Kb of acetate ion from Kw = Ka*Kb to check this out. It is less compared to, say, ammonia, which is a known weak base.
• at why did you not include Chlorine into the equation?
• See the chloride ion as the conjugate base of HCl, which is a very strong acid. Since a very strong acid has a very weak conjugate base, the chloride ion don't really take protons from water, thus it does not affect the pH of the solution, we thus do not include it into the equation. You may also refer to the previous video. This is similar to the reason why the chloride ion (and the sodium ion) in NaCl does not affect the pH of the solution.
• At ; how do you know which ions are going to react appreciably in water and thus use it in the equation?
• This is something you learn with experience, although it helps if you can remember the names of the common strong acids (HCl, HBr, Hi, H2SO4, HNO3, HClO4) and strong bases (hydroxides of Group 1 and 2 elements). There's a very good chance that if you have an acid or base that is not on this list, then it is a weak acid or base - this is particularly the case if it contains carbon (eg, CH3COOH). The conjugate bases of strong acids, and the conjugate acids of strong bases, do not react appreciably with water, whereas this is not the case with weak acids and bases.

In terms of the example you query, Cl- is the conjugate base of HCl. HCl is a very strong acid, which means that Cl- will be a very weak base. As a very weak base it will therefore not steal a proton from water to reform HCl.
• Why doesn't Na react with water? I thought H2O is polar and attracts Na?
• Metals like potassium and Sodium react violently with cold water. In case of sodium and potassium, the reaction is so violent and exothermic.
Sodium as an element is looking to donate one of its electrons to achieve a stable outer electron configuration. As an ion, sodium doesn't want to gain or lose any more electrons, so it isn't going to react with anything, including water.
• At , how can you always assume that the x is negligible? In what cases is it not? How can you make sure beforehand that you can just remove the x from the denominator like that?
• One "rule of thumb" that I learned is if your x ends up being larger than 5% of your starting value you need to solve the quadratic. Let's say that you started with an initial concentration of 5*10-8 M NH4Cl and you solve this problem using the method shown.
You will find x = 5.27*10-9
x is (5.27*10-9/5*10-8)*100% = 10.5% of starting value
If you solve the quadratic equation, you'll find x = 5.00*10-9

So, does this matter? Most of the time it doesn't BUT someday it might. So, in practice, it's worth checking if x is really negligible.
• What is the guarantee that CH3COONa will completely dissociate completely? I mean its also possible that only 0.15M dissociates. In that case answers would change. So are we to assume it dissociates completely??
• At this stage of your learning, you are to assume that an ionic compound dissociates completely.
And CH₃COO⁻Na⁺ is an ionic compound, so you assume that it dissociates completely.
• Do I create an ICE table on the MCAT or is there a more simple method to solve these problems? I'm specifically referring to the first example of the video. A lot of these examples require calculators and complex methods of solving.. help!