Inorganic Chemistry | - prove that pH=1/2[pKw+pKa+log C] -askIITians
1 Expert Answer(s) - - prove that pH=1/2[pKw+pKa+log C] for weak acid strong base salt solution. Answer this question and win exciting prizes. Acid Base Introduction · Acid Base Titration · Buffers And Hendersen · Conjugate Acids And Bases · Half Equivalence Point · PH Of A Weak Acid · PH Of A Weak. Im having trouble determining the relationship between pH and pKa and pKb. Is a low pKa representative of a low pH and a high pKb.
So this is equal to-- let me copy and paste this-- that. That's equal to that. So let's see if we can find a relationship between Ka and Kb. Well, one thing we can do is we can divide both sides by HA.
So if we divide both sides by HA. Actually, I could probably have that earlier on to the whole thing. If we ignore this part right here, this is equal to that. Let me erase all of this.
I'm using the wrong tool. So we could say that they both equal the concentration of A minus. So that's equal to that. We can divide both sides by HA. This over here will cancel with this over here.
And we're getting pretty close to a neat relationship. And so we get Ka over our hydrogen proton concentration is equal to our hydroxide concentration divided by Kb. You can just cross-multiply this. So we get Ka, our acidic equilibrium concentration, times Kb is equal to our hydrogen concentration times our hydroxide concentration. Remember, this is all in an aqueous solution.
What do we know about this? What do we know about our hydrogen times our hydroxide concentration in an aqueous solution? For example, let me review just to make sure I'm jogging your memory properly. We could have H2O. It can autoionize into H plus. And this has an equilibrium.
You just put the products. So the concentration of the hydrogen protons times the concentration of the hydroxide ions.
And you don't divide by this because it's the solvent. And we already figured out what this was. If we have just completely neutral water, this is 10 to the minus 7. And this is 10 to the minus 7. So this is equal to 10 to the minus Now, these two things could change.
I can add more hydrogen, I could add more hydroxide.
pKa and pKb relationship
And everything we've talked about so far, that's what we've been doing. That's what acids and bases do. They either increase this or they increase that. But the fact that this is an equilibrium constant means that, look, I don't care what you do to this. At the end of the day, this will adjust for your new reality of hydrogen protons.
And this will always be a constant.
As long as we're in an aqueous solution, a solution of water where water is a solvent at 25 degrees. I mean, in just pure water it's 10 to the minus 7. But no matter what we do to this and this in an aqueous solution, the product is always going to be 10 to the minus 14th power.
So that's the answer to this question. This is always going to be 10 to the minus If you multiply hydrogen concentration times OH concentration.
Now they won't each be 10 to the minus 7 anymore, because we're dealing with a weak acid or a weak base. So they're actually going to change these things. But when you multiply them, you're still going to get 10 to the minus And let's just take the minus log of both sides of that.
Let me erase all this stuff I did down here. I'll need the space. Let's say we take the minus logs of both sides of this equation. So you get the-- let me do a different color-- minus log, of course it's base 10, of Ka. Let me do it in the colors. Ka times Kb is going to be equal to the minus log of 10 to the minus So what is this equal to?
The log of 10 to the minus 14 is minus 14, because 10 to minus 14th power is equal to 10 to the minus The hormone-receptor complex then binds to DNA to effect transcription. Present in most cells, and is responsible for ACh-activated smooth muscle relaxation. Induced by cytokines to cause acute vasodilation. Forms free radical intermediates in PMN's and macrophages.
In the CNS, it is a Cl- channel. It binds at an allosteric site to increase the effectiveness of GABA.
The pKa Table Is Your Friend
It binds at a separate site than the barbiturates, but it is still GABAergic and binds at an allosteric site. Thus it is a convulsive agent. It binds excitatory neurotransmitters, glutamate and aspartate. But it doesn't work because it has a stimulatory effect on the hippocampus, causing hallucinations, similar to taking phencyclidine PCP. If you let ACh hang around long enough such in the presence of cholinesterase inhibitorsthen some of the ACh-receptors will convert to a high-affinity state, and the ACh will stay locked onto the receptors.
This explains the way in which cholinesterase inhibitors cause paralysis. Succinylcholine binds to the ACh with a higher affinity than ACh. Early on, you will see fasciculations, as it has its stimulatory effect on ACh. After that you see paralysis. Succinylcholine becomes an ACh antagonist, as all the receptors convert to the high-affinity state, and the molecule locks on.
A measure of the propensity of the drug to bind with a given receptor. A potent drug induces the same response at a lower concentration.
A potent drug has a lower EC50 value. The biologic response resulting from the binding of a drug to its receptor. An efficacious drug has a higher Emax value. A compound whose maximal response Emax is somewhat less than the full agonist. A plot of efficacy some measured value, such as blood pressure -vs- drug concentration.
The lower the EC50, the more potent the drug. A graph of discrete yes-or-no values, plotting the number of subjects attaining the condition such as death, or cure from disease -vs- drug concentration. The higher the therapeutic index, the better. So we're gonna make water here. And if NH four plus donates a proton, we're left with NH three, so ammonia.
Alright, let's think about our concentrations. So we just calculated that we have now. For ammonium, that would be. And for ammonia it was. So let's go ahead and write 0. So we're gonna lose all of this concentration here for hydroxide. And that's going to neutralize the same amount of ammonium over here. So we're gonna be left with, this would give us 0.18.2 Ka, pKa, Kb, pKb, Kw, pKw (HL)
Hydroxide we would have zero after it all reacts, And then the ammonium, since the ammonium turns into the ammonia, if we lose this much, we're going to gain the same concentration of ammonia.
So over here we put plus 0. So the final concentration of ammonia would be 0. And now we can use our Henderson-Hasselbalch equation. So let's go ahead and plug everything in. So ph is equal to the pKa. We already calculated the pKa to be 9.
And then plus, plus the log of the concentration of base, all right, that would be NH three. So the concentration of.
So this is all over. So if we do that math, let's go ahead and get out the calculator here and let's do this calculation. So let's get a little bit more room down here and we're done. The pH is equal to 9. So let's compare that to the pH we got in the previous problem.
The pKa Table Is Your Friend — Master Organic Chemistry
For the buffer solution just starting out it was 9. So we added a base and the pH went up a little bit, but a very, very small amount. So this shows you mathematically how a buffer solution resists drastic changes in the pH. Next we're gonna look at what happens when you add some acid. So we're still dealing with our same buffer solution with ammonia and ammonium, NH four plus. But this time, instead of adding base, we're gonna add acid.
And our goal is to calculate the pH of the final solution here. So the first thing we could do is calculate the concentration of HCl. So that would be moles over liters. That's our concentration of HCl. And HCl is a strong acid, so you could think about it as being H plus and Cl minus.