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I learned from many websites that $i$ of strong acids and bases and some weak bases could be determined by how many ions are dissociated. But, I realized that I need to calculate the $n$ in $i=1+(n-1)a$ ($a$ is the degree of dissociation), not the $i$ in my problems.

Is there any rule to calculate the $n$ for strong acids and bases and weak acids and bases? Do I also need to check solubility table to solve colligative property problems that using van't Hoff factor?

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  • $\begingroup$ Strong acids and bases dissociate completely. Why'd you put them in the same problem? And what's this with solubility? Of course precipitation changes the colligative properties. Please rethink and rephrase your question. $\endgroup$ – Karl Oct 3 '15 at 13:02
  • $\begingroup$ I didn't ask about that. I asked for the $n$ value. Please look my questions carefully. I saw that $/ce{H2SO4}$ has $i=3$, but is that the $n$ also 3? $\endgroup$ – làntèrn Oct 3 '15 at 13:09
  • $\begingroup$ The formula you cite is only useful if alpha is not one, i.e. a weakly dissociation compound. And n is the number of aggregates your original compound dissociates into. And no, HSO4- is a rather weak acid, so i is not 3 for sulfuric acid. $\endgroup$ – Karl Oct 3 '15 at 13:41
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In general, 'n' is the number of species the compound dissociates into, and such formulas are applicable only when the acid itself is monobasic, i.e.only one dissociaton is involved.For sulphuric acid,this is not applicable as there are two dissociations with two dissociation constants.The best way to find the van't hoff factor is to find the ratio of the final number of particles to the initial number of particles.

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You must know what van't hota factor is right. Its the ratio of observed colligative property and calculated or theoritical colligative property. And there are few more formulas like this.

n is the no. Of mole after dissociation or association what ever according to the problem.

Generally this formula is used to find degree of dissociation or association i.e, alpha.

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  • $\begingroup$ From what i have mentioned, i=3, but n=2 $\endgroup$ – shaistha Oct 3 '15 at 13:13
  • $\begingroup$ Did you mean any dissociation of any electrolyte, no matter weak or strong electrolyte? Will you put $n=3$ or $n=1$ for $\ce{H3PO4}$? $\endgroup$ – làntèrn Oct 3 '15 at 13:15
  • $\begingroup$ $\ce{H3PO4 <---> H+ + H2PO4-}$ is the equation for dissociationo $\endgroup$ – shaistha Oct 3 '15 at 13:24
  • $\begingroup$ Here both n and i are equal to 2 $\endgroup$ – shaistha Oct 3 '15 at 13:26
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    $\begingroup$ Why not 3H+ + (PO4)3- $\endgroup$ – làntèrn Oct 3 '15 at 13:30

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