Q4.21P - Chemistry: Molecular Nature Of Matter And Change

Question

Question: How many moles of  ions are present in the following aqueous solutions?

(a) 1.4 mL of 0.75 M hydrobromic acid 

(b) 2.47 mL of 1.98 M hydriodic acid 

(c) 395 mL of 0.270 M nitric acid

Step-by-Step Solution

Answer

Answer

 

The moles of H⁺ ions are present in the following aqueous solutions are:

(a)  $\mathbf{1}\mathbf{.}\mathbf{05}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\mathit{M}\mathit{o}\mathit{l}\mathbf{ }{\mathit{H}}^{\mathbf{+}}$

(b)  $\mathbf{4}\mathbf{.}\mathbf{89}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\mathit{M}\mathit{o}\mathit{l}\mathbf{ }{\mathit{H}}^{\mathbf{+}}$

(c) 0.107 mol  H⁺

Step 1: Calculate moles of H + ions released when hydrobromic acid dissolve in water

The dissociation reaction of hydrobromic acid is:

                     $\mathit{H}{\mathit{B}}_{\mathbf{r}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }{\mathit{H}}^{\mathbf{+}\mathbf{ }}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathit{B}}_{\mathbf{r}}\mathbf{-}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

The moles of${\mathit{H}}^{\mathbf{+}\mathbf{ }}$ is calculated as:   

$$\begin{gathered} \mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}{\mathit{H}}^{\mathbf{+}}\mathbf{=}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{ }\mathbf{ml}\mathbf{)}\mathbf{\left(}\frac{\mathbf{1}\mathbf{lit}}{\mathbf{1000}\mathbf{ml}}\mathbf{\left)}\mathbf{\right(}\frac{\mathbf{0}\mathbf{.}\mathbf{75}{\mathbf{molHB}}_{\mathbf{r}}}{\mathbf{Lit}}\mathbf{\left)}\mathbf{\right(}\frac{\mathbf{ }\mathbf{1}\mathbf{ }{\mathbf{molH}}^{\mathbf{+}}}{\mathbf{1}{\mathbf{molHB}}_{\mathbf{r}}}\mathbf{\right)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}\mathbf{ }}\mathit{m}\mathit{o}\mathit{l}{\mathit{H}}^{\mathbf{+}} \end{gathered}$$

 

Step 2: Calculate moles of ions released when hydriodic acid dissolve in water

 

The dissociation reaction of hydriodic acid is:

$\mathit{H}\mathit{I}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }{\mathit{H}}^{\mathbf{+}\mathbf{ }}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathit{I}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

     The moles of ${\mathit{H}}^{\mathbf{+}\mathbf{ }}$is calculated as:

Step 3: Calculate moles of ions released when nitric acid dissolve in water

 

The dissociation reaction of nitric acid is:

                   $\mathit{H}\mathit{N}{\mathit{O}}_{\mathbf{3}\mathbf{ }}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }{\mathit{H}}^{\mathbf{+}\mathbf{ }}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{+}\mathbf{N}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

The moles of ${\mathit{H}}^{\mathbf{+}}$ is calculated as:

$$\begin{gathered} \mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}{\mathit{H}}^{\mathbf{+}}\mathbf{=}\left(395 mi\right)\mathbf{\left(}\frac{\mathbf{1}\mathbf{l}\mathbf{i}\mathbf{t}}{\mathbf{1000}\mathbf{m}\mathbf{l}}\mathbf{\left)}\mathbf{\right(}\frac{\mathbf{0}\mathbf{.}\mathbf{270}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{H}\mathbf{N}{\mathbf{O}}_{\mathbf{3}}}{\mathbf{L}\mathbf{i}\mathbf{t}}\mathbf{\left)}\mathbf{\right(}\frac{\mathbf{ }\mathbf{1}\mathbf{ }\mathbf{m}\mathbf{o}\mathbf{l}{\mathbf{H}}^{\mathbf{+}}}{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{H}\mathbf{N}{\mathbf{O}}_{\mathbf{3}}}\mathbf{\right)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{107}\mathit{m}\mathit{o}\mathit{l}{\mathit{H}}^{\mathbf{+}} \end{gathered}$$