First year chem help (1 Viewer)

[Flop21]

This thread is for asking help in regards to any first year chem.

 

[Flop21]

So I did a back titration... with HCl and NaOH, and an unknown solid carbonate.

I know how to do the calculations, finding the mol of carbonate... but what is messing up my calculations is we diluted the initial carbonate/HCl solution. We put carbonate + 25mL HCL into a volumetric flask (100mL) and filled it up to the line with water. Then we use 10mL of this solution in the titration. So how do I factor this into my calculations??

 

[Flop21]

Pls help


So here's the question and answer, I don't understand what's going on in the answer at all.

I thought you could go, find moles of the acid and base, 0.013 mol and 0.006 mol respectively. Then go (0.013+0.006)/0.16L = 0.11875M.

Then to find pH, -log(0.11875) = 0.93.

Why is this wrong? What am I doing wrong here?

 

[marioyoshi]

Pls help


So here's the question and answer, I don't understand what's going on in the answer at all.

I thought you could go, find moles of the acid and base, 0.013 mol and 0.006 mol respectively. Then go (0.013+0.006)/0.16L = 0.11875M.

Then to find pH, -log(0.11875) = 0.93.

Why is this wrong? What am I doing wrong here?

The reason why that would be wrong is because then you are assuming acetic acid id a strong acid (ie degree os ionisation is 100%)
Thus using n = cv would not find the number of moles of H+ (which is relates to pH) but only the concentration of acid (which isnt related to pH)

 

[ml125]

Pls help


So here's the question and answer, I don't understand what's going on in the answer at all.

I thought you could go, find moles of the acid and base, 0.013 mol and 0.006 mol respectively. Then go (0.013+0.006)/0.16L = 0.11875M.

Then to find pH, -log(0.11875) = 0.93.

Why is this wrong? What am I doing wrong here?

You can't necessarily do that with this question, as acetic acid is not a strong acid - it does not completely dissociate in water. First, you need to find the moles of excess acid following the reaction, which would be 0.013-0.006=0.007. As shown in the answers you then use the ionisation constant (K_a) to find [H⁺] - as for any acid:

So, in this case:


The volumes will cancel, then rearrange to find[H⁺]. From the value of [H⁺] you get, you can find the pH.

If it were a strong acid, however, you would go (0.013-0.006)/0.16L = 0.04375. pH= -log(0.04375) =1.35

 

[Flop21]

How on earth am I suppose to know how and when to convert units.

E.g. when doing simple questions on photon energy and wavelength and such, they give you say 428 kJ mol^-1, but then in their working out they use it as 428x10^3. How do I know what to multiply it by? Some are x10^-9 etc.

 

[InteGrand]

How on earth am I suppose to know how and when to convert units.

E.g. when doing simple questions on photon energy and wavelength and such, they give you say 428 kJ mol^-1, but then in their working out they use it as 428x10^3. How do I know what to multiply it by? Some are x10^-9 etc.

Need to know the prefixes. k means 'kilo' (x10^3). A x10^{-9} would be nano.

 

[Flop21]

For electron configuration, so when you get up to the 4s orbital, do electrons then just also fill up 3d orbital? Instead of going 4s^2 3d^4 it's 4s^1 3d^5.

and why do some electron configurations skip 4s all together

 

[Flop21]

Need to know the prefixes. k means 'kilo' (x10^3). A x10^{-9} would be nano.

ohhhhhhhhhh

okay thanks will study these

 

[Flop21]

Why is BeCl2 not polar?

My understanding is the Be and Cl both have VERY different electronegativities... thus should be polar? Or is it non-polar because it's quite symmetrical?

 

[Flop21]

Calculating the concentration....


"calculate the concentration of KHP standad solution obtained by dissolving 5.10g of KHP into 250mL of water in a volumetric flask."


answers show: 1000/250 * 0.02497 (number of moles)

where did the 1000 come from?

 

[Flop21]

How does this simplify?


from ( P(Hg)^2 P(O2) ) / (HgO)^2

to this P(O2)

??

Like how does simplifying these sort of things work in chem.

 

[BlueGas]

Why is BeCl2 not polar?

My understanding is the Be and Cl both have VERY different electronegativities... thus should be polar? Or is it non-polar because it's quite symmetrical?

Determining whether a molecule is polar or not is pretty easy by knowing the overall dipole movement. Take a look at the picture below, the dipole vectors move towards the electronegative atom. If the overall dipole movement is 0, then it is non-polar. If the dipole vectors don't cancel each other out like PCl3, then it is polar. Since BeCl2 is linear, the dipole vectors cancel out and it is non-polar.

 

[BlueGas]

Calculating the concentration....


"calculate the concentration of KHP standad solution obtained by dissolving 5.10g of KHP into 250mL of water in a volumetric flask."


answers show: 1000/250 * 0.02497 (number of moles)

where did the 1000 come from?

n = m/mw, which you already found to be as 0.02497 moles.

C = n/v, where n is 0.02497 and v is 250ml/1000 to convert it to litres, and that'll get you the answer.

 

[strawberrye]

Calculating the concentration....


"calculate the concentration of KHP standad solution obtained by dissolving 5.10g of KHP into 250mL of water in a volumetric flask."

answers show: 1000/250 * 0.02497 (number of moles)

where did the 1000 come from?

Basically, concentration is moles/volume (L), in this instance 0.02497/(250ml in L)=0.02497/(250/1000), dividing by the latter fraction equals the same as times it by its reciprocal, hence you get 1000/250

 

[Flop21]

thanks


urgent question


when finding the conjugate acid/base for something... why do some things NOT have a conjugate acid/base??