R3.1.10 Calculating the Ka or Kb of a weak acid or base (HL)

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Mike Sugiyama Jones

5 жыл бұрын

This video covers how to calculate the Ka or Kb of a weak acid or base.

Пікірлер: 13

@mohammedhamouda1607 5 жыл бұрын

You are doing great work!! I am just wondering is there parts of the syllabus that you still did not cover?

@MSJChem 5 жыл бұрын

I would say that I’ve covered 99% of the syllabus - they may be parts that I haven’t covered but I couldn’t tell you what they are.

@aht3261 3 жыл бұрын

how did you get the 14.00 value in pOH sir?

@MSJChem 3 жыл бұрын

At 298 K, pH + pOH = 14.00

@hanamehmeti3216 10 ай бұрын

1:05 where does the [H+] = 5.89 x 10-5 mol dm-3 came from?

@MSJChem 10 ай бұрын

10^-4.23 (10 raised to the power of -4.23)

@beepbobeep4594 3 ай бұрын

Dear Mr Mike, At the risk of sounding dumb, could I please ask why we always need equilibrium concentrations for these equations. Why couldn't we just use 0.1 mol dm3 and the concentration of [H+] without the ICE process. Thank you so much for your help

@MSJChem 3 ай бұрын

Because acid dissociate into ions so we need to know the [H+]. If we make a few assumptions we don’t need to make an ICE box.

@beepbobeep4594 3 ай бұрын

@@MSJChem Thank you so much! Does that mean that if we’re not dealing with weak acids/bases, we have to do the ICE box, but if we are dealing with weak acids and bases, we assume that the concentration at equilibrium is not significantly different to the starting concentration and therefore we don’t need the ICE box

@MSJChem 3 ай бұрын

@beepbobeep4594 for strong acids and bases there is complete dissociation so no need for an ICE box. For weak acids and bases there is also no need if we make the assumption.

@beepbobeep4594 3 ай бұрын

@@MSJChem Oh right! Of course, because strong acids/bases are not at an equilibrium due to complete dissociation. Thank you Mr Mike!