Question

How do you calculate the pH of a ${\text{0}}{\text{.26M}}$ methylamine solution?

Answer 1

For the calculation of the pH of a methylamine solution, first we have to calculate the ${\text{pOH}}$ of the given compound because it is basic in nature, and after getting the value of ${\text{pOH}}$ we will calculate the value of pH by the following equation for an aqueous solution: ${\text{pH + pOH = 14}}$.

Complete step by step answer:
- As we know that methyl amine is a weak base and its dissociation reaction is written as:
${\text{C}}{{\text{H}}_{\text{3}}}{\text{N}}{{\text{H}}_{{\text{2}}\left( {{\text{aq}}} \right)}}{\text{ + }}{{\text{H}}_{\text{2}}}{{\text{O}}_{\left( {\text{l}} \right)}} \rightleftharpoons {\text{C}}{{\text{H}}_{\text{3}}}{\text{NH}}_{\text{3}}^{\text{ + }}{\text{ + O}}{{\text{H}}^{\text{ - }}}$
- And from the above reaction it is clear that ${\text{x}}$ moles of hydroxide ion (${\text{O}}{{\text{H}}^{\text{ - }}}$) and ${\text{C}}{{\text{H}}_{\text{3}}}{\text{NH}}_{\text{3}}^{\text{ + }}$ will form from the ${\text{0}}{\text{.26M}}$ of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{N}}{{\text{H}}_3}$.
-Equilibrium constant or base dissociation constant equation for above given equation is written as:
${{\text{K}}_{\text{b}}}{\text{ = }}\dfrac{{\left\lceil {{\text{O}}{{\text{H}}^{\text{ - }}}} \right\rceil \left[ {{\text{C}}{{\text{H}}_{\text{3}}}{\text{NH}}_{\text{3}}^{\text{ + }}} \right]}}{{\left[ {{\text{C}}{{\text{H}}_{\text{3}}}{\text{N}}{{\text{H}}_{\text{3}}}} \right]}}{\text{ = }}\dfrac{{{\text{x}}{\text{.x}}}}{{{\text{2}}{\text{.6 - x}}}}$
And according to the table of ${{\text{K}}_{\text{b}}}$ which is given in internet, value of equilibrium constant for methylamine is $4.4 \times {10^{ - 4}}$.
Hence, $4.4 \times {10^{ - 4}}{\text{ = }}\dfrac{{{\text{x}}{\text{.x}}}}{{{\text{2}}{\text{.6 - x}}}}$
$4.4 \times {10^{ - 4}}\left( {{\text{2}}{\text{.6 - x}}} \right){\text{ = }}{{\text{x}}^2}$
Or ${{\text{x}}^{\text{2}}}{\text{ - 11}}{\text{.44 \times 1}}{{\text{0}}^{{\text{ - 4}}}}{\text{ + 4}}{\text{.4x = 0}}$
After solving the above equation we can get the value of concentration in terms of moles of hydroxide ion (${\text{O}}{{\text{H}}^{\text{ - }}}$) and ${\text{C}}{{\text{H}}_{\text{3}}}{\text{NH}}_{\text{3}}^{\text{ + }}$ i.e. which is equal to${\text{0}}{\text{.0105mol/L}}$.
-Now from the concentration of hydroxide ion (${\text{O}}{{\text{H}}^{\text{ - }}}$), value of ${\text{pOH}}$ will be calculated as follow:
${\text{pOH = - lo}}{{\text{g}}_{10}}\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]$
${\text{pOH = - lo}}{{\text{g}}_{10}}\left[ {0.0105} \right]$
Or ${\text{pOH = - lo}}{{\text{g}}_{10}}\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right] = 1.98$
-For an aqueous solution relation between pH and ${\text{pOH}}$ will be written as:
${\text{pH + pOH = 14}}$
- On putting the value of ${\text{pOH}}$ on the above equation, we will find the value of pH for the given ${\text{0}}{\text{.26M}}$ solution of methylamine.
${\text{pH + 1}}{\text{.98 = 14}}$
Or ${\text{pH = 12}}{\text{.02}} \simeq {\text{12}}$
Hence, the pH of a ${\text{0}}{\text{.26M}}$ methylamine solution approximately equals $12$.

Note: Here some of you may think that in the equilibrium constant equation why we put minus ${\text{x}}$ in the denominator for the concentration of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{N}}{{\text{H}}_3}$, so the reason is that at equilibrium condition ${\text{x}}$ moles of product will be formed from the given moles of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{N}}{{\text{H}}_3}$ that’s why we put minus ${\text{x}}$ there.