Topics 8.1 - 8.3

[@mrfarabaugh]

2:07:30 Oops! That Kb expression is written correctly, but it should be labeled as Kb (not Ka). bit.ly/3z7Zp6v

[@exoisbased]

My chemistry teacher shared this after meeting you at ACS Spring, and honestly these explanations are great 😊

[@kaitlyn8185]

At 1:08:49, I was hoping you could clarify the difference between partially ionized and partially soluble. For example, if a weak acid like HF is highly soluble, does that mean that H2O molecules can break apart the intermolecular attractions between multiple HF molecules so it dissolves? Thank you!!

[@mrfarabaugh]

Think of it like this. Some substances such as sugar or ethanol are very soluble in water, but they are nonelectrolytes. They mix with water molecules very well to form a homogeneous solution, but they don’t break up into ions. Some substances such as NaCl or KNO₃ are very soluble in water and also strong electrolytes. They mix with water molecules very well to form a homogeneous solution and they also will break up completely into positive and negative ions. Then there are weak acids and weak bases such as vinegar and ammonia.They also mix with water very well. But they ionize to a small extent. Since they are only partially ionized in solution, they are weak electrolytes. So the ability to form a homogeneous solution is one property that is unrelated to a substance’s ability to break up or ionize in the solution. With sugar or ethanol, they can form hydrogen bonds with water. With ionic substances, they can form ion-dipole forces with water. With weak acids and bases it can be a combination of these attractive forces.

[@Srvelis82]

Excited about this one. I use these videos as end of unit study guides. Students are to complete the packet to the best of their ability, then watch the video to make corrections.

[@mrfarabaugh]

Thank you. Unit 8 can be challenging for students. I hope that your students find this video helpful.

[@kaitlyn8185]

At 1:52:07, shouldn't the conjugate base of a weak acid be a strong conjugate base? Because if the acid is weak, it doesn't want to give up its H+ ions, and instead wants to accept them?

[@mrfarabaugh]

According to the AP Chemistry curriculum, there are six STRONG ACIDS that students should memorize: HCl, HBr, HI, HClO₄, HNO₃, and H₂SO₄. These acids have rather large Ka values and are 100% ionized in aqueous solution. The six conjugate bases of these six strong acids (chloride, bromide, iodide, perchlorate, nitrate, and hydrogen sulfate) are so weak, that they are considered as negligible bases, and don't affect the pH of the solution. Examples of STRONG BASES are metal hydroxides in which the metal is located in Groups 1 & 2, such as NaOH, KOH, Ca(OH)₂, and Sr(OH)₂. When these ionic compounds dissolve in water, they split up to produce cations and hydroxide ions. The conjugate acid of the hydroxide ion is water. Everything else that is classified as an acid or a base tends to fall into the WEAK category. I consider acetic acid to be a weak acid (Ka = 1.8E-5) and the acetate ion (Kb = 5.6E-10) to be a weak base. I consider ammonia to be a weak base (Kb = 1.8E-5) and the ammonium ion (Ka = 5.6E-10) to be a weak acid. When I compare two different weak acids or two different weak bases by looking at their relative K values, I use words like stronger and weaker as comparative words. Acetic acid is a stronger acid than the ammonium ion (but they are both classified as weak acids). Ammonia is a stronger base than the acetate ion (but they are both classified as weak bases).

[@kaitlyn8185]

​@@mrfarabaugh Ohh I see, so I should use the Ka or Kb, and the acids and bases AP wants us to know to determine the relative strength of the acid or base? Thank you for taking the time to reply!!

[@teckinblade]

how does 1.8x10^-5 / 0.5 not equal 3.6x10^-5? am i missing something? my calculator disagrees with this video, i am confused 1:24:56

[@mrfarabaugh]

To solve that equation on the right side of the screen, you need to multiply both sides of the equation by 0.50. 'This gives the following equation: x² = (1.8×10⁻⁵)(0.50) = 9.0×10⁻⁶

[@teckinblade]

oh i see now that the exponent changed to make the expression true, thank you for responding!