Respuesta :
ANSWER
The pH value is 4.88
EXPLANATION
Given information
[tex]\begin{gathered} \text{ The concentration of HCN is 0.27M} \\ \text{ The K}_a\text{ value of HCN is 6.2 }\times\text{ 10}^{-10} \end{gathered}[/tex]To find the pH of the concentration of the solution, follow the steps below
Step 1: Write the ionic equation of the reaction.
[tex]\text{ CN}^-+\text{ H}_2O\rightleftarrows\text{ HCN + OH}^-[/tex]Step 2: write the chemical equilibrium of the reaction
[tex]\begin{gathered} \text{ CN}^-+\text{ H}_2O\text{ }\rightleftarrows\text{ HCN + OH}^- \\ \text{ Initial conc. 0.27 } \\ \text{ Change in conc. - x +x} \\ \text{ Equilibrium \lparen0.27 - x\rparen x} \\ \\ \text{ K}_a\text{ = }\frac{[HCN]}{[CN^-]\frac{}{}\text{ / \lbrack CN}^-]} \end{gathered}[/tex][tex]\begin{gathered} \text{ Recall, that K}_{a\text{ }}=\text{ 6.2}\times10^{-10} \\ 6.2\times10^{-10\text{ }}=\text{ }\frac{x^2}{(0.27\text{ - x\rparen}} \\ \text{ since \lbrack HCN\rbrack/K}_a\text{ is > 100, we can write the equation above as} \\ 6.2\text{ }\times\text{ 10}^{-10}\text{ = }\frac{x^2}{0.27} \\ \text{ cross multiply} \\ 6.2\text{ }\times10^{-10}\text{ }\times0.27\text{ = x}^2 \\ 1.674\text{ }\times\text{ 10}^{-10}\text{ = x}^2 \\ \text{ Take the square roots of both sides} \\ \sqrt{1.674\times10^{-10}}=\text{ x} \\ 1.294\text{ }\times\text{ 10}^{-5\text{ }}\text{ = x} \\ \text{ Hence, 1.294 }\times\text{ 10}^{-5\text{ }}\text{ = \lbrack HCN\rbrack} \end{gathered}[/tex]Step 3: Find the pH of the solution using the below formula
[tex]\begin{gathered} \text{ pH = -log \lbrack H}^+\text{\rbrack} \\ \text{ pH = -log \lbrack1.294}\times10^{-5}] \\ pH\text{ = 4.88} \end{gathered}[/tex]Hence, the pH of the solution is 4.88
