A student makes up a solution of propan-2-amine in water with a concentration of 1.00 mol L ¯1.
- Using structural formulae, complete the equation for the reaction of propan-2-amine with water. (2 marks)
- The equilibrium constant for the reaction of propan-2-amine with water is `4.37 xx10^(-4)`.
- Calculate the concentration of hydroxide ions in this solution. (3 marks)
--- 5 WORK AREA LINES (style=lined) ---
a.
b.
\begin{array} {|l|c|c|c|}
\hline & \ce{C3H7NH2} & \ce{C3H7NH3} & \ce{OH–} \\
\hline \text{Initial} & 1.00 & 0 & 0 \\
\hline \text{Change} & -x & +x & +x \\
\hline \text{Equilibrium} & 1.00-x & x & x \\
\hline \end{array}
\[ K_b = \ce{\frac{[C3H7NH3+][OH– ]}{[C3H7NH2 ]}} = \frac{x^2}{(1.00-x)} \]
Assume `1.00-x=1.00` because `x` is negligible:
| `4.37 xx 10^(−4)` | `= x^2 / 1.00` | |
| `x` | `=sqrt(4.37 xx 10^(−4))` | |
| `= 0.0209\ text{mol L}^(–1)` |
`=> [text{OH}^– ] = 0.0209\ text{mol L}^(–1)`
a.
b.
\begin{array} {|l|c|c|c|}
\hline & \ce{C3H7NH2} & \ce{C3H7NH3} & \ce{OH–} \\
\hline \text{Initial} & 1.00 & 0 & 0 \\
\hline \text{Change} & -x & +x & +x \\
\hline \text{Equilibrium} & 1.00-x & x & x \\
\hline \end{array}
\[ K_b = \ce{\frac{[C3H7NH3+][OH– ]}{[C3H7NH2 ]}} = \frac{x^2}{(1.00-x)} \]
Assume `1.00-x=1.00` because `x` is negligible:
| `4.37 xx 10^(−4)` | `= x^2 / 1.00` | |
| `x` | `=sqrt(4.37 xx 10^(−4))` | |
| `= 0.0209\ text{mol L}^(–1)` |
`=> [text{OH}^– ] = 0.0209\ text{mol L}^(–1)`
