How do you find concentration of ions? I found the Concentration of the solution now what?

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How do you find concentration of ions? I found the Concentration of the solution now what?

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Answer 1 · 2017-11-21T18:47:19

Well, are there any data here...?

Explanation:

Let us suppose that you dissolved a mass of $71.02 \cdot g$ $71.02*g$ of $N a_{2} S O_{4} (a q)$ $Na_2SO_4(aq)$ , a soluble salt, in a $1 \cdot L$ $1*L$ volume of water....

And thus............... $Concentration of sodium sulfate = \frac{Moles of solute}{Volume of solution}$ $"Concentration of sodium sulfate"="Moles of solute"/"Volume of solution"$

$[N a_{2} S O_{4}] \equiv \frac{\frac{71.02 \cdot g}{142.04 \cdot g \cdot m o l^{- 1}}}{1 \cdot L} = 0.500 \cdot m o l \cdot L^{- 1}$ $[Na_2SO_4]-=((71.02*g)/(142.04*g*mol^-1))/(1*L)=0.500*mol*L^-1$

Now in SOLUTION this salt undergoes ionization to give THREE EQUIV of ions.....i.e.

$N a_{2} S O_{4} (s) H_{2} O - - \to 2 N a^{+} + S O_{4}^{2 -}$ $Na_2SO_4(s) stackrel(H_2O)rarr2Na^+ +SO_4^(2-)$ ...and given this speciation...we got....

$[N a^{+}] = 2 \times 0.500 \cdot m o l \cdot L^{- 1} = 1.00 \cdot m o l \cdot L^{- 1}$ $[Na^+]=2xx0.500*mol*L^-1=1.00*mol*L^-1$ ...

$[S O_{4}^{2 -}] = 1 \times 0.500 \cdot m o l \cdot L^{- 1} = 0.500 \cdot m o l \cdot L^{- 1}$ $[SO_4^(2-)]=1xx0.500*mol*L^-1=0.500*mol*L^-1$ ...

And so, you have to work out how the salt reacts in solution; to which ions does it speciate? Does this address your question?

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How do you find concentration of ions? I found the Concentration of the solution now what?

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