Q4.20P - Chemistry: Molecular Nature Of Matter And Change

Question

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

(a) 1.40 L of 0.25 M perchloric acid 

(b) 6.8 mL of 0.92 M nitric acid 

(c) 2.6 L of 0.085 M hydrochloric acid

Step-by-Step Solution

Answer

Answer

 

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

(a) 0.35 mol${\mathit{H}}^{\mathbf{+}}$

(b)$\mathbf{6}\mathbf{.}\mathbf{26}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}\mathbf{ }}\mathit{m}\mathit{o}\mathit{l}\mathbf{ }{\mathit{H}}^{\mathbf{+}}$

(c) 0.221 mol ${\mathit{H}}^{\mathbf{+}}$

 

Step 1: Calculate moles of H + ions released when HCIO 4 dissolve in water

 

The dissociation reaction of is:

         $\mathit{H}\mathit{C}\mathit{I}{\mathit{O}}_{\mathbf{4}\mathbf{ }}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }{\mathit{H}}^{\mathbf{+}\mathbf{ }}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathit{C}\mathit{I}{{\mathit{O}}^{\mathbf{-}}}_4\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$                      

The moles of H⁺ is calculated as:   

$$\begin{gathered} \mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}{\mathit{H}}^{\mathbf{+}}\mathbf{=}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{40}\mathbf{ }\mathit{L}\mathbf{)}\mathbf{\left(}\frac{\mathbf{1}\mathbf{lit}}{\mathbf{1000}\mathbf{ml}}\mathbf{\left)}\mathbf{\right(}\frac{\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{ }{\mathbf{molHCIO}}_4}{\mathbf{Lit}}\mathbf{\left)}\mathbf{\right(}\frac{\mathbf{ }\mathbf{1}\mathbf{ }{\mathbf{molH}}^{\mathbf{+}}}{\mathbf{1}\mathbf{mol}\mathbf{ }{\mathbf{HCIO}}_{\mathbf{4}}}\mathbf{\right)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{35}\mathbf{ }\mathit{m}\mathit{o}\mathit{l}{\mathit{H}}^{\mathbf{+}} \end{gathered}$$

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

 

The dissociation reaction of HN0₃ is:

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

The moles of H⁺ is calculated as:   

$$\begin{gathered} \mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}{\mathit{H}}^{\mathbf{+}}\mathbf{=}\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{8}\mathbf{ }\mathbf{mI}\mathbf{)}\mathbf{\left(}\frac{\mathbf{1}\mathbf{lit}}{\mathbf{1000}\mathbf{ml}}\mathbf{\left)}\mathbf{\right(}\frac{\mathbf{0}\mathbf{.}\mathbf{92}\mathbf{ }{\mathbf{molHNO}}_{\mathbf{3}}}{\mathbf{Lit}}\mathbf{\left)}\mathbf{\right(}\frac{\mathbf{ }\mathbf{1}\mathbf{ }{\mathbf{molH}}^{\mathbf{+}}}{\mathbf{1}{\mathbf{molHNO}}_{\mathbf{3}}}\mathbf{\right)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{26}\mathbf{ }\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\mathit{m}\mathit{o}\mathit{l}{\mathit{H}}^{\mathbf{+}} \end{gathered}$$

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

 

The dissociation reaction of HCl is:

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

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

$$\begin{gathered} \mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}{\mathit{H}}^{\mathbf{+}}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{6}\mathbf{ }\mathit{L}\mathbf{)}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{85}\mathbf{ }\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{ }\mathbf{H}\mathbf{C}\mathbf{I}}{\mathbf{LIt}}\mathbf{\left)}\mathbf{\right(}\frac{\mathbf{ }\mathbf{1}\mathbf{ }{\mathbf{molH}}^{\mathbf{+}}}{\mathbf{1}\mathbf{molHCI}}\mathbf{\right)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{221}\mathit{m}\mathit{o}\mathit{l}{\mathit{H}}^{\mathbf{+}} \end{gathered}$$