Question

Problem:Morphine is a weak base for which\({{\bf{K}}_{\bf{b}}}\)is 8 ×\({\bf{1}}{{\bf{0}}^{{\bf{ - 7}}}}\). Calculatethe pH of a solution made by dissolving 0.0400 molof morphine in water and diluting to 600.0 mL.

Short answer

The pH of the morphine solution is 10.87.

Step-by-step solution

Given data

The mole of the pentafluorobenzoic acid is 0.0400 mol.

The\({{\rm{K}}_{\rm{b}}}\)of morphine base is\(8 \times {10^{ - 7}}\).

The volume of the solution is 600.0mL which in litre in 0.6 L.

Chemical reaction

The chemical reaction when the morhphine baseis dissolved in water that results in the formation of the morhphinebaicsolution is written as follows:

\({{\bf{C}}_{{\bf{17}}}}{{\bf{H}}_{{\bf{19}}}}{{\bf{O}}_{\bf{3}}}{\bf{N + }}{{\bf{H}}_{\bf{2}}}{\bf{O}} \to {{\bf{C}}_{{\bf{17}}}}{{\bf{H}}_{{\bf{19}}}}{{\bf{O}}_{\bf{3}}}{\bf{N}}{{\bf{H}}^{\bf{ + }}}{\bf{ + O}}{{\bf{H}}^ - }\)

The molarity of the morphine basic solution is as follows:

\(\begin{aligned}{}{\bf{M = }}\dfrac{{{\bf{mole}}\;{\bf{(n)}}}}{{{\bf{Volume}}\;{\bf{in}}\;{\bf{L}}}}\\{\rm{ = }}\;\dfrac{{{\rm{0}}{\rm{.400}}\;{\rm{mol}}}}{{0.6\;{\rm{L}}}}\\{\rm{ = }}\;{\rm{0}}{\rm{.66M}}\end{aligned}\)

calculating the pH

The hydroxide ion concentration is calculated as follows:

\(\begin{aligned}{}{\bf{O}}{{\bf{H}}^{\bf{ - }}}{\bf{ = }}\sqrt {\left( {{{\bf{C}}_{{\bf{17}}}}{{\bf{H}}_{{\bf{19}}}}{{\bf{O}}_{\bf{3}}}{\bf{N}}} \right){\bf{ \times }}{{\bf{K}}_{\bf{b}}}} \\{\rm{ = }}\;\sqrt {{\rm{0}}{\rm{.66 \times 8}} \times {\rm{1}}{{\rm{0}}^{ - 7}}} \\{\rm{ = }}\;0.0007261\;\end{aligned}\)

The pH of morphineis calculated using the formula as follows:

\(\begin{aligned}{}{\bf{pH = 14 + log}}\left( {{\bf{O}}{{\bf{H}}^{\bf{ - }}}} \right)\\ = 14 + \log \left( {0.0007261} \right)\\ = 14 + \left( { - 3.13} \right)\\ = 10.87\end{aligned}\)

Therefore, the pH of the morphine solution is 10.87.