| Date | November 2014 | Marks available | 1 | Reference code | 14N.1.hl.TZ0.27 |
| Level | HL | Paper | 1 | Time zone | TZ0 |
| Command term | Question number | 27 | Adapted from | N/A |
Question
Methylamine acts as a weak base when it reacts with water. For a diluted aqueous solution, what is the \({K_{\text{b}}}\) expression for this reaction?
A. \({K_{\text{b}}} = \frac{{{\text{[C}}{{\text{H}}_{\text{3}}}{\text{NH}}_{\text{3}}^ + {\text{][O}}{{\text{H}}^ - }{\text{]}}}}{{{\text{[C}}{{\text{H}}_{\text{3}}}{\text{N}}{{\text{H}}_{\text{2}}}{\text{]}}}}\)
B. \({K_{\text{b}}} = \frac{{{\text{[C}}{{\text{H}}_{\text{3}}}{\text{N}}{{\text{H}}_{\text{2}}}{\text{][}}{{\text{H}}_{\text{2}}}{\text{O]}}}}{{{\text{[C}}{{\text{H}}_{\text{3}}}{\text{NH}}_{\text{3}}^ + {\text{][O}}{{\text{H}}^ - }{\text{]}}}}\)
C. \({K_{\text{b}}} = \frac{{{\text{[C}}{{\text{H}}_{\text{3}}}{\text{NH}}_{\text{3}}^ + {\text{][O}}{{\text{H}}^ - }{\text{]}}}}{{{\text{[C}}{{\text{H}}_{\text{3}}}{\text{N}}{{\text{H}}_{\text{2}}}{\text{][}}{{\text{H}}_{\text{2}}}{\text{O]}}}}\)
D. \({K_{\text{b}}} = \frac{{{\text{[C}}{{\text{H}}_{\text{3}}}{\text{N}}{{\text{H}}_{\text{2}}}{\text{]}}}}{{{\text{[C}}{{\text{H}}_{\text{3}}}{\text{NH}}_{\text{3}}^ + {\text{][O}}{{\text{H}}^ - }{\text{]}}}}\)
Markscheme
A
Examiners report
Quite a few candidates included water in the expression, giving answer C.
