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CHEMISTRY, M6 2025 HSC 11 MC

The structures of two substances, \(\text{X}\) and \(\text{Y}\), are shown.
 

Which row of the table correctly classifies these substances as a Brønsted-Lowry acid or a Brønsted-Lowry base?
 

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ & \\
 \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Brønsted-Lowry} & \textit{Brønsted-Lowry} \\
\textit{acid}\rule[-1ex]{0pt}{0pt}& \textit{base} \\
\hline
\rule{0pt}{2.5ex}\text{-}\rule[-1ex]{0pt}{0pt}&\text{X and Y}\\
\hline
\rule{0pt}{2.5ex}\text{X and Y}\rule[-1ex]{0pt}{0pt}& \text{-}\\
\hline
\rule{0pt}{2.5ex}\text{Y}\rule[-1ex]{0pt}{0pt}& \text{X} \\
\hline
\rule{0pt}{2.5ex}\text{X}\rule[-1ex]{0pt}{0pt}& \text{Y} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution
  • A Brønsted-Lowry acid is a proton donor and a Brønsted-Lowry base is a proton acceptor.
  • \(X\) is propanoate (the conjugate base of propanoic acid) and is therefore a proton accepter making it a Brønsted-Lowry base.
  • \(Y\) is ethanamine and is considered a weak base where the \(\ce{NH2}\) group can accept a proton to become \(\ce{NH3+}\).

\(\Rightarrow A\)

CHEMISTRY, M5 2020 HSC 27

A student makes up a solution of propan-2-amine in water with a concentration of 1.00 mol L ¯1.

  1. Using structural formulae, complete the equation for the reaction of propan-2-amine with water.   (2 marks)
     
         
  2. The equilibrium constant for the reaction of propan-2-amine with water is  `4.37 xx10^(-4)`.
  3. Calculate the concentration of hydroxide ions in this solution.   (3 marks)

Show Answers Only

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)`

Show Worked Solution

a.   

   


♦ Mean mark (a) 48%.

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)`


Mean mark (b) 51%.