Problem: Calculate the concentration of all species present and the pH of a 0.020-M HF solution.

🤓 Based on our data, we think this question is relevant for Professor Halpin's class at NYU.

FREE Expert Solution

Equilibrium reaction:        HF(aq) + H2O ⇌ F-(aq) + H3O+

Step 1: Construct an ICE chart for the reaction.

Step 2: Write the Ka expression.

${\mathbf{K}}_{{\mathbf{a}}}\mathbf{=}\frac{\mathbf{products}}{\mathbf{reactants}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{a}}\mathbf{=}\frac{\mathbf{\left[}{\mathbf{F}}^{\mathbf{-}}\mathbf{\right]}\mathbf{\left[}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\right]}}{\left[\mathrm{HF}\right]}$

Ka of HF = 7.2x10-4

Step 3: Calculate the equilibrium concentrations.

${\mathbf{K}}_{\mathbf{a}}\mathbf{=}\frac{\left[\mathbf{x}\right]\left[\mathbf{x}\right]}{\left[\mathbf{0}\mathbf{.}\mathbf{020}\mathbf{-}\mathbf{x}\right]}\phantom{\rule{0ex}{0ex}}\mathbf{7}\mathbf{.}\mathbf{2}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{0}\mathbf{.}\mathbf{020}\mathbf{-}\mathbf{x}}$

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$\mathbf{7}\mathbf{.}\mathbf{2}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{\left(}\mathbf{0}\mathbf{.}\mathbf{020}\mathbf{-}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}}{\overline{)\mathbf{0}\mathbf{.}\mathbf{020}\mathbf{-}\mathbf{x}}}\overline{)\left(0.020-x\right)}\phantom{\rule{0ex}{0ex}}\mathbf{1}\mathbf{.}\mathbf{44}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{-}\mathbf{7}\mathbf{.}\mathbf{2}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{x}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\phantom{\rule{0ex}{0ex}}\overline{)\mathbf{0}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{7}\mathbf{.}\mathbf{2}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{44}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}}$

Now, we will use the quadratic formula to solve for x. The quadratic formula is:

$\overline{){\mathbf{x}}{\mathbf{=}}\frac{\mathbf{-}\mathbf{b}\mathbf{±}\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{a}\mathbf{c}}}{\mathbf{2}\mathbf{a}}}$

$\mathbf{x}\mathbf{=}\frac{\mathbf{-}\mathbf{7}\mathbf{.}\mathbf{2}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{±}\sqrt{\mathbf{\left(}\mathbf{7}\mathbf{.}\mathbf{2}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}{\mathbf{\right)}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{\left(}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{44}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{\right)}}}{\mathbf{2}\mathbf{\left(}\mathbf{1}\mathbf{\right)}}\phantom{\rule{0ex}{0ex}}\mathbf{x}\mathbf{=}\frac{\mathbf{-}\mathbf{7}\mathbf{.}\mathbf{2}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{±}\sqrt{\mathbf{5}\mathbf{.}\mathbf{184}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{76}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}}}{\mathbf{2}}\phantom{\rule{0ex}{0ex}}\mathbf{x}\mathbf{=}\frac{\mathbf{-}\mathbf{7}\mathbf{.}\mathbf{2}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{±}\sqrt{\mathbf{5}\mathbf{.}\mathbf{81184}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}}}{\mathbf{2}}\phantom{\rule{0ex}{0ex}}\mathbf{x}\mathbf{=}\frac{\mathbf{-}\mathbf{7}\mathbf{.}\mathbf{2}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{±}\mathbf{7}\mathbf{.}\mathbf{62}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}{\mathbf{2}}$

The calculation then splits into two ways:

Problem Details

Calculate the concentration of all species present and the pH of a 0.020-M HF solution.