Q4.53P - Chemistry: Molecular Nature Of Matter And Change

Question

An auto mechanic spills 88 mL of 2.6 M H₂SO₄ solution from a rebuilt auto battery. How many milliliters of 1.6 M NaHCO₃ must be poured on the spill to react completely with the sulfuric acid?

Step-by-Step Solution

Answer

286 ml of 1.6 M NaHCO₃ must be poured on the spill to react completely with the sulfuric acid.

Step 1:first write down the balanced molecular equation:

$\mathbf{2}\mathit{N}\mathit{a}\mathit{H}\mathit{C}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{S}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{\to }\mathit{N}{\mathit{a}}_{\mathbf{2}}\mathit{S}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathit{H}}_{\mathbf{2}}\mathit{C}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}$

Step 2: To find out the volume of NaHCO 3 poured to react completely with the sulfuric acid, first, we have to calculate the moles of NaHCO 3 :

Given values are: 

 $$\begin{gathered} volumeofstandard2.6M{H}_{2}S{O}_{4}solution=88ml \\ volumeofNaHC{O}_3=1.6M \end{gathered}$$

Moles of NaHCO₃ is: 

 $\begin{array}{rcl}\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathit{o}\mathit{f}\mathit{N}\mathit{a}\mathit{H}\mathit{C}{\mathit{O}}_{\mathbf{3}}& \mathbf{=}& \mathbf{\left(}\mathbf{88}\mathit{m}\mathit{l}{\mathit{H}}_{\mathbf{2}}\mathit{S}{\mathit{O}}_{\mathbf{4}}\mathbf{\right)}\mathbf{\left(}\frac{\mathbf{1}\mathbf{l}\mathbf{i}\mathbf{t}}{\mathbf{1000}\mathbf{m}\mathbf{l}}\mathbf{\right)}\mathbf{\left(}\frac{\mathbf{2}\mathbf{.}\mathbf{6}\mathbf{m}\mathbf{o}\mathbf{l}{\mathbf{H}}_{\mathbf{2}}\mathbf{S}{\mathbf{O}}_{\mathbf{4}}}{\mathbf{1}\mathbf{l}\mathbf{i}\mathbf{t}{\mathbf{H}}_{\mathbf{2}}\mathbf{S}{\mathbf{O}}_{\mathbf{4}}\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}}\mathbf{\right)}\mathbf{\left(}\frac{\mathbf{2}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{N}\mathbf{a}\mathbf{H}\mathbf{C}{\mathbf{O}}_{\mathbf{3}}}{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}{\mathbf{H}}_{\mathbf{2}}\mathbf{S}{\mathbf{O}}_{\mathbf{4}}}\mathbf{\right)}\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{4576}\mathit{m}\mathit{o}\mathit{l}\mathit{N}\mathit{a}\mathit{H}\mathit{C}{\mathit{O}}_{\mathbf{3}}\\ & & \end{array}$

The volume of NaHCO₃ poured to react completely with the sulfuric acid:

 $\mathit{v}\mathit{o}\mathit{l}\mathit{u}\mathit{m}\mathit{e}\mathit{o}\mathit{f}\mathit{N}\mathit{a}\mathit{H}\mathit{C}{\mathit{O}}_{\mathbf{3}}\mathbf{=}\frac{\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{s}\mathbf{o}\mathbf{f}\mathbf{N}\mathbf{a}\mathbf{H}\mathbf{C}{\mathbf{O}}_{\mathbf{3}}}{\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{i}\mathbf{t}\mathbf{y}\mathbf{o}\mathbf{f}\mathbf{N}\mathbf{a}\mathbf{H}\mathbf{C}{\mathbf{O}}_{\mathbf{3}}}$

$\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{4576}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{N}\mathbf{a}\mathbf{H}\mathbf{C}{\mathbf{O}}_{\mathbf{3}}}{\mathbf{\left(}\frac{\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{N}\mathbf{a}\mathbf{H}\mathbf{C}{\mathbf{O}}_{\mathbf{3}}}{\mathbf{1}\mathbf{l}\mathbf{i}\mathbf{t}\mathbf{N}\mathbf{a}\mathbf{H}\mathbf{C}{\mathbf{O}}_{\mathbf{3}}}\mathbf{\right)}}$

$\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{286}\mathit{l}\mathit{i}\mathit{t}\mathit{N}\mathit{a}\mathit{H}\mathit{C}{\mathit{O}}_{\mathbf{3}}\mathbf{=}\mathbf{286}\mathit{m}\mathit{l}\mathit{N}\mathit{a}\mathit{H}\mathit{C}{\mathit{O}}_{\mathbf{3}}$