Learn about Senior Chemistry: Acids and Bases 12 Video

Learn about Senior Chemistry, Acids and Bases 12, in this comprehensive video by bannanaiscool.
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In this question, we’re not seeking in an unknown concentration of volume. We have the concentration in volume of both chemicals we mix together we just want to know what the pH is of this solution in the end. So when we take these two chemicals and mix them together, we may not have the same number of moles to be able to get total neutralization or an equivalent number of moles. So what do we need to do? We need to find the moles of each and see who might be in excess. So, here’s the reaction. We have acetic acid reacting with the NaOH forms acetate ion and water and here’s the concentration in volume given of acetic acid concentration of volume given of the NaOH. So, we can take moles per liters and find moles of this. Moles per liter and five moles of that. Note that these two chemicals should react to the one to one ratio to form products. If we find the moles of the acetic acid by just multiplying the concentration in moles per liter times the liters, we get this millimole. We do the same for the NaOH and here’s the story, look at this. We have present ready to react 0.0015 moles of acetic acid, but we have more NaOH. We have 0.00200 moles of NaOH. So then this reacts with this, how much of the NaOH reacts? Well if it’s a 1:1 and it is, then 0.0015 here only reacts with 0.0015 there. I'm subtracting this from this and now it’s here. But you know what I'm doing. I'm really saying is the difference between these two now because that will tell you how many moles of NaOH is left un-reacted. When the you take away these two numbers from each other you get 0.0005 moles of NaOH left over. So now if you can find the concentration of that chemical left over, you’ve got the pH like this. So, we have the moles in excess, but is the number of moles of OH left over the reaction. The new volume of the solution is what you divided into the moles to get the moles per litter. The volume of the OH ion that is now in solution is not the original 10ml of NaOH and not the 15 ml of the acetic acid with the combination of the two because you’d mix them together. The new volume of the solution is 25 ml divide that ml into that mmol to get 0.020 moles/liter NaOH. How do you get the pH? It’s 14 plus the OH ion concentration and you get 12.30. So that’s in excess of the base. It is the base that is in more quantity in the solution in the end. What if the acid was actually in excess? Let’s do one of those.

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