# A 5.00*g mass of "sodium bicarbonate" requires WHAT volume of 3.00*mol*L^-1 HCl for stoichiometric equivalence?

Aug 17, 2017

We need (i) a stoichiometric equation.....and (ii) we need approx. $20 \cdot m L$ of the acid solution.

#### Explanation:

$N a H C {O}_{3} \left(a q\right) + H C l \left(a q\right) \rightarrow N a C l \left(a q\right) + C {O}_{2} \left(g\right) \uparrow + {H}_{2} O \left(l\right)$

And thus (ii) there is 1:1 molar equivalence between hydrochloric acid and sodium bicarbonate.

Of course we are not done. We have to work out the number of moles of sodium bicarbonate and hydrochloric acid.

We know that $\text{concentration"="moles of stuff"/"volume of solution}$

And we know that $\text{moles of stuff"="mass of stuff"/"molar mass of stuff}$.

So....$\text{Moles of sodium bicarbonate} = \frac{5.00 \cdot g}{84.0 \cdot g \cdot m o {l}^{-} 1}$

$= 0.0595 \cdot m o l$

And given the equation, I need an equivalent quantity of $H C l \left(a q\right)$.

i.e. $\text{Volume of HCl} = \frac{0.0595 \cdot m o l}{3.00 \cdot m o l \cdot {\mathrm{dm}}^{-} 3} = 0.0198 \cdot {\mathrm{dm}}^{3}$.

Since there are $1000 \cdot m L \cdot {\mathrm{dm}}^{-} 3$, we need approx. $20 \cdot m L$.

This does seem a lot of work. But the answer depends on the two defining equations, i.e. $\text{concentration}$, and $\text{moles}$. Of course a stoichiometrically balanced equation was crucial.

Just on the relationship between ${\mathrm{dm}}^{3}$, and the more familiar $L$ (at least it is more familiar in the States!).

We know that $1 \cdot {\mathrm{dm}}^{3} \equiv {\left(1 \times {10}^{-} 1 \cdot m\right)}^{3} = {10}^{-} 3 \cdot {m}^{3}$; $d$ means $\text{deci}$, i.e. a ${10}^{-} 1$ multiplier. But there are ${10}^{3} \cdot L \cdot {m}^{-} 3$. Thus $1 \cdot {\mathrm{dm}}^{3} = {10}^{-} 3 \cdot {m}^{3} \equiv 1 \cdot L$ as required.