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Cell diagram .txt | And electrons travel from this guy to this guy. |
Cell diagram .txt | So the way you read this is that copper solid oxidize releasing two electrons and this ion. |
Cell diagram .txt | Now, these two electrons travel to this side into the cathode. |
Cell diagram .txt | And in the cathode, they react with the cadmium, forming our cadmium solid. |
Cell diagram .txt | And this is how you read the cell diagram for any electrochemical cell. |
Half equivalence point .txt | In this lecture, we're going to talk about Titrating, an asset using a base. |
Half equivalence point .txt | Now, if you don't know what Titration is and you don't know what an equivalence point Is, then check out the link below. |
Half equivalence point .txt | So, once again, we're tightrading an acid with the base. |
Half equivalence point .txt | And here's our Titration curve where the y axis is PH and the x axis is the volume of base added. |
Half equivalence point .txt | Now, we've defined the equivalence point to be the point at which all the acid has been neutralized. |
Half equivalence point .txt | So every single molecule of acid in our buffer system has been neutralized. |
Half equivalence point .txt | Now, we can also define something called the half equivalence point. |
Half equivalence point .txt | And the half equivalence point is the point at which exactly half of the acid in our buffer system has been neutralized. |
Half equivalence point .txt | Now, suppose we choose our buffer system to consist of acetic acid. |
Half equivalence point .txt | Now, acetic acid associates it to acetate ion and an H plus ion. |
Half equivalence point .txt | Now, suppose we begin adding some volume of our base. |
Half equivalence point .txt | And suppose we choose our base to be ammonia. |
Half equivalence point .txt | So we're adding the volume of ammonia into our acetic acid. |
Half equivalence point .txt | Now, the reaction looks like this. |
Half equivalence point .txt | The conjugate acid, acetic acid dissociates into the conjugate base acetate ion. |
Half equivalence point .txt | This base, the ammonia gains an H becoming ammonia. |
Half equivalence point .txt | Now, so, let's look at this definition again. |
Half equivalence point .txt | The half equivalence point is the point at which exactly half of the assets, exactly half of this guy has dissociated into this guy Aka. |
Half equivalence point .txt | Has been neutralized. |
Half equivalence point .txt | So that means we can define the half equivalence point in another way. |
Half equivalence point .txt | The half equivalence point is the point at which the concentration of the conjugate acid equals the concentration of the conjugate base. |
Half equivalence point .txt | Because half of this is now this. |
Half equivalence point .txt | So the concentration of this guy equals this guy. |
Half equivalence point .txt | Well, why is this definition important? |
Half equivalence point .txt | Well, we'll see why in a second. |
Half equivalence point .txt | Let's look at the Henderson Hasselblack formula or equation. |
Half equivalence point .txt | Now, if you don't know what this formula is, check out the link above. |
Half equivalence point .txt | So, this equation states that PH is equal to PKA of our acid plus log of this ratio the concentration of the conjugate base over the concentration of the conjugate acid. |
Half equivalence point .txt | And this PH is the PH of our buffer system. |
Half equivalence point .txt | So notice that since this guy equals this guy, this divided by this is one. |
Half equivalence point .txt | So what's inside here is simply one. |
Half equivalence point .txt | So let's rewrite it. |
Half equivalence point .txt | PH is equal to PKA plus log of one. |
Half equivalence point .txt | But what's log of one? |
Half equivalence point .txt | Well, log of one is zero. |
Half equivalence point .txt | And that means PH equals PKA. |
Half equivalence point .txt | Well, that's nice and all, but why is that important? |
Half equivalence point .txt | Where's the significance? |
Half equivalence point .txt | Well, this means that now we can choose the PH of our bumper system by simply choosing an acid with PKA that's closest to our desired PH. |
Half equivalence point .txt | So suppose, for example, I want my bubble system to have a PH of 4.7. |
Half equivalence point .txt | Now, how I find the asset to use is I simply find the acid with the PKA value closest to 4.7. |
Half equivalence point .txt | Now, I go online I find my table, I look up an acid with a PH that's a PKA closest to a PH of 4.7, and I find that it's acetic acid. |
Half equivalence point .txt | So now I know, using this equation here, that if I choose my buffer sydney system to consist of acetic acetic acid, my PH of my bumper system will be 4.7. |
Half equivalence point .txt | And that's important. |
Parts per million Example .txt | Parts per million is a way of finding the concentration of the solution. |
Parts per million Example .txt | It's represented by the letters Ppm or parts per million. |
Parts per million Example .txt | The formula is mass of compound x divided by total mass of solution multiplied by ten to the six or million. |
Parts per million Example .txt | Now, since this is a ratio that units cancel and Ppm is unitless, so now it's still an example using parts per million. |
Parts per million Example .txt | In this example, we start with 25 bowls of water in a cup. |
Parts per million Example .txt | Here's our cup. |
Parts per million Example .txt | The blue dots are the water molecules, nothing else exists. |
Parts per million Example .txt | You want to find them out in grams of HCL to add to create a 90,000 parts per million solution. |
Parts per million Example .txt | So you want to go from this cup to this cup, where this cup contains HCL molecules in the concentration of 90,000 parts per million. |
Parts per million Example .txt | We want to find the amount of grams of the red dots we need to add to create such a solution. |
Parts per million Example .txt | The first step is to calculate the molecular weight of water. |
Parts per million Example .txt | To calculate the molecular weight of water, we simply add the atomic weight of oxygen plus two times the atomic weight of H because we have a substrate of two, so we get 16 grams/mol of oxygen plus two times 1 gram/mol of H gives us 18 grams/mol. |
Parts per million Example .txt | So the molecular weight of water is 18 grams/mol. |
Parts per million Example .txt | Now, to find the amount in grams of water that we have in our initial solution, we need to multiply the molecular weight by the 25 moles of H 20 that we have. |
Parts per million Example .txt | So 18 grams/mol times 25 moles gives you 450. |
Parts per million Example .txt | Now, moles cancel, the grams are left, so we have 450 grams of H 20. |
Parts per million Example .txt | Now, we want to find the amount of HCL we need to add in terms of grams to create a 90,000 parts per million solution. |
Parts per million Example .txt | So we simply use the parts per million formula. |
Parts per million Example .txt | We say, well, we want to create a 90,000 grams or parts per million solution equals x is the amount of HCL grams we need to add divided by the total amount of grounds we have the solution. |
Parts per million Example .txt | So we already have 450 grams of water plus the amount of HCL and grounds we will add. |
Parts per million Example .txt | So plus x times ten to the 6th or 1 million. |
Parts per million Example .txt | We do a little bit of simple algebra to solve for x, we divide through by ten to the six we get 90,000 divided by 1 million equals this guy. |
Parts per million Example .txt | Now, we multiply through by 450 plus x and we get 0.9, which is simply this guy times 450 plus x equals the x was left over on that side, so equals x. |
Parts per million Example .txt | We saw for x you get x equals 44.51 grams of HCL. |
Parts per million Example .txt | So we want to add 44 grams, 44.51
grams of HCL to our initial 450 grams of water to to create a 90,000 parts per million solution of HCL. |
Solubility Product Constant .txt | In this lecture, we're going to talk about the solubility of powder constant, KSP. |
Solubility Product Constant .txt | Before we talk about KSP, let's talk about the solubility of ionic compounds. |
Solubility Product Constant .txt | Now, all ionic compounds have the ability to dissociate into their ion form when added into water. |
Solubility Product Constant .txt | For example, let's take ionic compound sodium chloride. |
Solubility Product Constant .txt | When we add sodium chloride chloride into water, it dissociates into two ions, sodium and chloride. |
Solubility Product Constant .txt | Now, this reaction is called the forward reaction or dissolution. |
Solubility Product Constant .txt | The reverse reaction is just as likely to occur, and that's called precipitation, the sufformation of ionic compound from its ion form. |
Solubility Product Constant .txt | Now, initially, when we add photos chloride into water, the forward rate is much higher than the reverse rate. |
Solubility Product Constant .txt | Eventually, however, though dynamic equilibrium is achieved, at this point, the forward rate is equal to the reverse rate. |
Solubility Product Constant .txt | And at this point, the solution is said to be saturated, which basically means that the concentration of the ions or dissolved ions is that it's maximum. |
Solubility Product Constant .txt | So these guys are at the maximum. |
Solubility Product Constant .txt | Now, whenever we talk about normal equations or normal reactions, not fluidation reactions, we talk about equilibrium constants. |
Solubility Product Constant .txt | In the same way, when we talk about salvation reactions, we could talk about something called solubility product constant or KST. |
Solubility Product Constant .txt | Now, when we determine the normal equilibrium constant, we don't include solids and liquids in our calculation. |
Solubility Product Constant .txt | And in the same way, when we talk about salvation or solubility product constant KFC, we don't include solids and liquids. |
Solubility Product Constant .txt | For example, let's take a reaction of solid Orion compound AB that associates in water into A plus B. |
Solubility Product Constant .txt | Now, since we don't count the solids, we don't count the liquids, but we do count gases and Aqueous compounds. |
Solubility Product Constant .txt | When we determine the KSP or the solubility of bonus constants, we don't count this guy or the other guy. |
Solubility Product Constant .txt | We only count these two guys. |
Solubility Product Constant .txt | So KSP is equal to the concentration of A times the concentration of B. |
Solubility Product Constant .txt | In this problem, we're given some unknown amount of barium sulfate and some unknown amount of water in a cup. |
Solubility Product Constant .txt | Now, we want to mix the two and wait for dynamic equilibrium to establish. |
Solubility Product Constant .txt | Once equilibrium establishes, we're given that the KSP or the Solubility product is equal to 1.0 times ten to negative ten at 25 degrees Celsius. |
Solubility Product Constant .txt | So we want to find the solubility of barium sulfate. |
Solubility Product Constant .txt | To find the solubility of barium sulfate, we must first write the dissociation reaction for barium sulfate. |
Solubility Product Constant .txt | Therefore, we get 1 mol of barium sulfate in its solid form, dissociates into 1 mol of barium plus 1 mol of sulfate, and both guys are in the Aqueous form. |