1 ml of $H_3PO_4$ was diluted to $250$ mL. $25$ mL of this solution required $40$ mL of $0.1$ N $NaOH$ for neutralisation. The density of the acid is $\dots$?

@Nick Thank you.

Here's my attempt: milliequivalents of NaOH = 0.1 * 40 = 4 = milliequivalents of H3PO4 = 25 * Normality of H3PO4. So the normality of H3PO4 = 0.16. I'm not sure what I'm supposed to do with this information.

@ParthKohli: You have an equation now. Can you try to figure it out from that

@ManishEarth: Hiya. How are you?

OK, we'll do it your way. 40 mL of 0.1 N NaOH would be 4 millimoles, so we've got 4/3 millimoles of H3PO4.

The thing is that I want to understand it using equivalents rather than moles.

Think...

Yeah, hold on.

Did my attempt assume the correct path?

To get from point A to point B, you must know where point B is.

Hence, what are you searching for?

I'm searching for the density of the acid. In other words, I'm searching for its mass in 1 ml.

The 1 ml was diluted to 250 ml and the concentration of 250 ml is 0.16 N.

Would that mean that the concentration of pure H3PO4 is (0.16 * 250) N?

Ah, hmm.

Great, so the concentration of pure H3PO4 is 40 N. Its molarity is 40/3, in that case, and so 40/3 moles of H3PO4 are there in 1 liter.

40/3 moles => mass = 40/3 * (3 + 31 + 64) = 40/3 * 98 g = 1306.666... g

1.306 kg/liter

Express it in $g/cm^3$

Oh.

1306 g/1000 cc = 1.306 g/cc

Yeah that looks fine.

@ParthKohli: Now, I suggest you do these questions

Stoichiometry is fun :D Don't you think?

I guess...

Hmm, could you help me with one more thing?

Not stoichiometry this time though.

kay :D

(Hopefully not equillibria)

I've still not gotten the hang of oxidation numbers/states. I know how to calculate it and whatnot, but they still haven't made peace with me.

If I've been given the structure of, say the tetrathionate ion for instance, how do I calculate the oxidation number of each atom?

$S_4O_6^{2−}$

What is the oxidation number of O?

-2 of course.

And that of sulphur here

(that's the question :D)

Actually, wait a second. This would correctly give me the average oxidation state.

But it differs for each atom.

Two atoms have +5, two have 0.

Yeah, it's weird in this case.

@Nick o/

Hey, does anyone here live in/near Portland?

Usually we just do $4S + 6(-2) = -2$ and be done with it.

@Nick Of course, that's what.

@ManishEarth: Heya :D Can you help us figure out the mystery of this ion?

?

sure

Okay,

..and?

The more I read about oxidation states, the more I get confused about what they are.

@ParthKohli: Maybe Hank Green can explain it you better than we can.

(Sorry now, I gotta run.)

@ParthKohli usually we assume that the S-S bonds cause no oxidation number

@ManishEarth Why?

Is that true for all elements?

Usually there's no polarization along such bonds

I'm not sure if this happens to all cases though

you could have a highly polar element-element bond provided the other groups attached are appropriately polarizing

Could this perhaps help any? chemistry.stackexchange.com/a/4837

The process to determine oxidation numbers is almost the same as for formal charges

The only difference is that for formal charges you cleave all bonds in the middle, while for oxidation numbers you drag both bond electrons to the more electronegative atom

Ah!

Thank you. :)