Question

Calculate the pH of a 0.27 M HCN solution. Ka(HCN) = 6.2x10^-10Give your answer to 2 places after the decimal.

Answer 1

The pH value is 4.88

Step-by-step explanation

Given information

`\begin{gathered} \text{ The concentration of HCN is 0.27M} \\ \text{ The K}_a\text{ value of HCN is 6.2 }*\text{ 10}^(-10) \end{gathered}`

To find the pH of the concentration of the solution, follow the steps below

Step 1: Write the ionic equation of the reaction.

`\text{ CN}^-+\text{ H}_2O\rightleftarrows\text{ HCN + OH}^-`

Step 2: write the chemical equilibrium of the reaction

`\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}`
`\begin{gathered} \text{ Recall, that K}_{a\text{ }}=\text{ 6.2}*10^(-10) \\ 6.2*10^{-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{ }*\text{ 10}^(-10)\text{ = }(x^2)/(0.27) \\ \text{ cross multiply} \\ 6.2\text{ }*10^(-10)\text{ }*0.27\text{ = x}^2 \\ 1.674\text{ }*\text{ 10}^(-10)\text{ = x}^2 \\ \text{ Take the square roots of both sides} \\ \sqrt{1.674*10^(-10)}=\text{ x} \\ 1.294\text{ }*\text{ 10}^{-5\text{ }}\text{ = x} \\ \text{ Hence, 1.294 }*\text{ 10}^{-5\text{ }}\text{ = \lbrack HCN\rbrack} \end{gathered}`

Step 3: Find the pH of the solution using the below formula

`\begin{gathered} \text{ pH = -log \lbrack H}^+\text{\rbrack} \\ \text{ pH = -log \lbrack1.294}*10^(-5)] \\ pH\text{ = 4.88} \end{gathered}`

Hence, the pH of the solution is 4.88