## Nominal Concentration

504608290
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Joined: Fri Sep 25, 2015 3:00 am

### Nominal Concentration

For number 12.25, it asks you to calculate the initial molarity of $Ba(OH)_{2}$ and the molarities of $Ba^{^{2+}}$, $OH^{^{-}}$, and $H_{3}O^{^{+}}$ in an aqueous solution that contains .43 g of $Ba(OH)_{2}$ in .1L of solution. How do you do this problem? In the solutions manual it mentions that $[Ba(OH)_{2}]_{0}$ = nominal concentration of $Ba(OH)_{2}$. What does this mean? I've searched everywhere and the term nominal concentration doesn't seem to exist.

Annie Qing 2F
Posts: 28
Joined: Fri Sep 25, 2015 3:00 am

### Re: Nominal Concentration

So the dissociation reaction is $Ba(OH)_{2}\rightarrow Ba^{2+}+2OH^{-}$

First, calculate the initial molarity of $Ba(OH)_{2}$:
$[Ba(OH)_{2}]=\left ( \frac{0.43 g Ba(OH)_{2}}{0.100 L} \right )\left ( \frac{mol Ba(OH)_{2}}{171.342 g Ba(OH)_{2}} \right )=0.025 M Ba(OH)_{2}$
where 171.342 g/mol is the molar mass of $Ba(OH)_{2}$.

Then, since we know $Ba(OH)_{2}$ completely dissociates, by the dissociation reaction above:
$[Ba^{2+}]$ = 0.025 M
$[OH^{-}]$ = 2(0.025 M) = 0.050 M

Finally, since we know $[OH^{-}]$ = 0.050 M from above,
$[H_{3}O] = \frac{K_{w}}{\left [ OH^{-} \right ]}=\frac{1.0x10^{-14}}{0.050 M}=2.0x10^{-13}M$

I think the solutions manual was simply trying to clarify that putting brackets around a chemical formula denotes the concentration of that molecule or compound.

Hopefully this helps!

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