JEE CHEMISTRY MCQ: Equilibrium

1 Medium

1 of 1

Application +4 | -1

Morphine

\left(\mathrm{C}_{17} \mathrm{H}_{19} \mathrm{NO}_3\right)

is administered medically to relieve pain. It is a naturally occurring base or alkaloid. What is the

\mathrm{pH}

of a

0.0025 \mathrm{~M}

solution of morphine at

25^{\circ} \mathrm{C}?

The base-ionization constant

K_{{b}}

1.6 \times 10^{-6}

25^{\circ} \mathrm{C}.

(\log 2=0.3)

View answer and explanation

Correct option: B

9.8

Option B (9.8) is correct because the pH is calculated from the pOH using the base ionization constant.

More Questions from Equilibrium

Practice All Chapter MCQs →

\mathrm{K}_1

\mathrm{K}_2

\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\quad\ldots

\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\quad\ldots

K{a_1},\,\,\,K{a_2}\,\,\,\mathrm{and}\,\,\,K{a_3}

\begin{array}{l} {H_2}S\leftrightharpoons{H^ + } + H{S^ - }\\ H{S^ - }\leftrightharpoons{H^ + } + {S^{2 - }}\\ {H_2}S\leftrightharpoons2{H^ + } + {S^{2 - }} \end{array}

K{a_1},\,\,K{a_2}\,,K{a_3}

\mathrm{pH}

0.1 \mathrm{~M}-\mathrm{N}_2 \mathrm{H}_4

\mathrm{N}_2 \mathrm{H}_4,

K_{{b}_1}=3.6 \times 10^{-6}, K_{{b}_2}=6.4 \times 10^{-12}, \log 2=0.3

\log 3=0.48

{\text{Ba}}{{\text{O}}_2}(s) \rightleftharpoons {\text{BaO}}(s) + {{\text{O}}_2}(g);\Delta H = - {\text{ve}}.

\text{O}_2