Question

How many grams of solid sodium cyanide should be added to 1.00 L of a 0.119 M hydrocyanic acid solution to prepare a buffer with a pH of 8.809

Answer 1

Answer:

1.62 g

Explanation:

Given that:

Concentration of HCN = 0.119 M

Assuming the ka 4.00 × 10⁻¹⁰

The pKa of  HCN (hydrocyanic acid)  = -log (Ka)

= - log ( 4.00 × 10⁻¹⁰)

= 9.398

pH of buffer = 8.809

Using Henderson Hasselbach equation:

$pH = pKa + log \dfrac{[conjugate\ base ]}{acid}$

$pH = pKa + log \dfrac{[CN^-]}{[HCN]}$

$8.809 = 9.398 +log \dfrac{[CN^-]}{[HCN]}$

$log \dfrac{[CN^-]}{[HCN]}= 8.809 - 9.398$

$log \dfrac{[CN^-]}{[HCN]}= -0.589$

$\dfrac{[CN^-]}{[HCN]}= 0.2576$

[CN^-] = 0.2576[HCN]

[CN^-] = 0.2756 (0.119) L

[CN^-] = 0.033 M

∴

The amount of NaCN (sodium cyanide) is calculated as follows:

$= 1.00 L \times \dfrac{0.033 \ mol \ NacN }{1 \ L } \times \dfrac{49.01 \ g}{1 \ mol \ of \ NacN}$

= 1.62 g