Recent content by ollerkandi

ok Then forget it, I will just find someone else who is much more competent in explaining. Goodbye.

You only told me this is wrong, but I still do 't know how to appropriately apply the mols to the equations that you put. I just need that to understand the bigger picture of how to use the mols in the reaction.

why can't you just tell me the full solution? This is not a homework or anything (I don't know why the mod moved this thread here), I just need to understand how it works. Like in the easier example, there is 0.005 CH3COOH and reacting with 0.0051 CH3COONA. It leaves 0.0001 mol of CH3COONA and...

So does the coefficient matter? Because in the original equation, there is 2HCL, but in the one you posted it's only CO3 + H --> HCO3. Does that mean that there will be 0,0816 mol of HCO3? and then 0,0816 of H2CO3? Assuming volume is 1 L, does this mean that the pH is pH = pKa - log...

ok I give up.. I really don't understand why it works differently with CH3COONA + CH3COOH where 0,005 mol CH3COONA + 0,0051 CH3COOH --> 0,0001 CH3COOH + 0,005 CH3COONA (straightforward reaction)

I'm following the stoichiometry, and that's why I still can't understand how to solve the problem. This is the original equation NA2CO3 + 2HCL --> 2NACL + H2CO3 We have 0,244 mol NA2CO3 neutralizing 0,0816 HCL. From what I understand (and I'm following how the teacher solves the problem in the...

(0.224 x 2) - 0.0816, I suppose

H2CO3...

It's not enough I suppose.. but then why is there an acid in the solution?

thank you for the reply and sorry for the confusing format.. so does it mean that in the solution we have (0,224 mol x 2 minus 0,0816 mol CO3- and 0,0816 mol H2CO3? Sorry if I haven't understood the full picture.

Hello, I just had my first exam yesterday, and I'm always tripped with this question that combines neutralization with buffer solution. The task is to find the pH. I did try to understand the general concept behind it. From my understanding, if CH3COOH reacts with NAOH and the acid has a higher...