Question

Calculate the pH of a 0.437 M aqueous solution of hydrocyanic acid (HCN, Ka = 4.0×10-10) and the equilibrium concentrations of the weak acid and its conjugate base.

Answer 1

pH=4.88

`[HCN]_(eq)=0.43699M`

`[CN^-]_(eq)=1.322x10^(-5)M`

Step-by-step explanation:

Hello,

In this case, the undergoing dissociation reaction is:

`HCN(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+CN^-(aq)`

In such a way, the law of mass action becomes:

`Ka=([H^+]_(eq)[CN^-]_(eq))/([HCN]_(eq))`

Which in terms of the change
`x` due to the reaction extent, goes:

`Ka=((x)(x))/(0.437-x)=4.0x10^(-10)`

Thus, solving for
`x` by either quadratic equation or solver, the results are:

`x_1=1.322x10^(-5)M\\x_2=-1.322x10^(-5)M`

Clearly, the answer is:

`x_1=1.322x10^(-5)M`

In this manner, since
`x` equals the concentration of hydrogen ions, the pH turns out:

`pH=-log([H^+])=-log(1.322x10^(-5))=4.88`

And the concentration of the HCN and the CN⁻:

`[HCN]_(eq)=0.437M-1.322x10^(-5)M=0.43699M`

`[CN^-]_(eq)=1.322x10^(-5)M`

Best regards.