Question #f4f3c

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Question #f4f3c

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Answer 1 · 2017-09-04T16:17:08

(a.) $0.25 M H C l$ $0.25M HCl$
(b.) $0.50 M H C l$ $0.50M HCl$
(c.) $0.20 M N a_{3} P$ $0.20M Na_3P$

Explanation:

This question is focusing on the van't Hoff factor $i$ $i$ , the amount of particles the solute breaks up into. This can be experimentally determined, or for the purpose of most general chemistry questions like this one, estimate to the nearest integer, as such:

a. $i_{H C l} = 2$ $i_(HCl) = 2$ , since $H C l$ $HCl$ will dissociate into $H^{+}$ $H^+$ and $C l^{-}$ $Cl^-$ ions.

Thus,

$0.20 M \cdot 2 = 0.40 M$ $0.20M * 2 = 0.40M$ v. $0.25 M \cdot 2 = 0.50 M$ $0.25M * 2 = 0.50M$

A similar method can be used to solve (b.) and (c.). Acetic acid will break up into $H^{+}$ $H^+$ and $C_{3} H_{2} O_{3}^{-}$ $C_3H_2O_3^-$ ions; $N a_{3} P$ $Na_3P$ will break up into 3 $N a^{+}$ $Na^+$ ions and one $P^{-}$ $P^-$ ion.

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Question #f4f3c

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