Question

Question: How many moles ofions 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

Short 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-by-step solution

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:

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

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:

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:

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