Recent content by njh

Thankyou.

Thank you. I believe that to be the ICE process.

Thankyou Borek, very helpful. There was a gap in my understanding, which is now more clear.

I think the answer is that in a balanced equation you do have the fixed molarity. If there is an excess concentration of one of the Reactants, then this remains unutilised.

Thank you again for the clarification. For what should have been a simple concept, I feel that I made a three course meal of it; for which I appreciate people's effort.

Thank you again. I was confusing the concepts of Protonation and Dissociation. I have looked again at protonation with reference to titration of an amino acid. I understand that "Protonation" refers to the extent that a molecule has an [H+] available to donate. In the case of COOH ↔ COO-...

Thank you, I understand now. Once an acid has reached equilibrium in a solution, it can not donate more Hydrogen ions, even if the concentration of the acid is increased. This is because there is nothing to accept these Hydrogen ions. As a result, the concentration of undissociated acid will...

Thank you for the question. At a lower pH, the original acid donates more Hydrogen ions (which I think is due to a difference in electronegativity). So the acid is the protonator.

I understand the HH equation as: ##pH = pKa + log\frac{[A^-]}{[HA]}## The Numerator of the fraction is the conjugate Base of the original acid. The denominator is the undissociated/ unprotonated Acid. I understand that when the pH equals the PkA, then the solution is 50% dissociated, so...

Fantastic, so -2/3. There were a couple of things that I hit a wall on. The first was watching the negatives at each stage, which with hindsight just requires care. The second was that I initially tried to get rid of the square root as a way to simplify, which proved wrong. The one thing that...

I think bamboum just meant in terms of whether the numerator or denominator dominates the equation, so they are just saying that the denominator dominates, which as a starting indicator sounds correct. Oh, but I think that I see what you mean. Whether it is + or - is not.

Yes, that is the correct function. Thank you for the notation correction, as well as the reminder on addition and subtraction under square roots, which I see would have led me down the wrong path. I see the subsequent pointer from WWGD, so I'll try and answer that next and see where it takes me...

##\frac{\sqrt{16x^6}-\sqrt{x^2}}{6x^3 + x^2}## ##\frac{4x^3-\sqrt{x^2}}{6x^3+x^2}## ##\frac{4-\sqrt{x^2}}{6+x^2}## My request is may I confirm that I have this correct up to this point? I do know the final answer, I know the suggested process for calculating the answer, but I want to check...

Thank you (and incidentally, my apologies for not using mathematical notation in my question).

I understand the mathematics that 1 divided by infinity is virtually zero and so equals zero. I look on the internet and that is the answer that I get. Is this a simplification for early mathematics learning and, if I continue, will I find a more complex answer? The reason that I ask is that I...