The acid-dissociation constant of The acid-dissociation constant of hydrocyanic acid (HCN) at 25.0°C is 49x 10-10. You have an aqueous solution of 0.080 M sodium cyanide (NaCN). What type of salt is this? circle one acidic basic neutral D) 8.8 What is the pH of 0.080 M sodium cyanide (NaCN) ? A) 5.2 "B) 2.89 C) 11.11.

1. basic

2. C

Explanation:

1. CN⁻ is the conjugate base of HCN, so the salt is basic.

2. The net balanced chemical equation is:

CN⁻ + H₂O ⇒ HCN + OH⁻

The acid-dissociation constant Kb for this reaction can be found from the Ka for the conjugate acid:

Kb = 10⁻¹⁴/Ka = 10⁻¹⁴ / (4.9 x 10¹⁰) = 2.040816 ... x10⁻⁵

Kb is the equilibrium constant for the above reaction and is expressed in terms of the equilibrium concentrations of the aqueous species in the reaction as follows:

Kb = [HCN][OH⁻ ]/[CN⁻]

The CN⁻ in the solution dissociates by amount "x" into "x" amount of HCN and OH⁻. Therefore, expressions for the equilibrium concentrations of these species can be substituted into the expressions for Kb

Kb = 2.040816 ... x10⁻⁵ = [HCN][OH⁻ ]/[CN⁻] = (x) (x) / (0.080 - x)

The result is a quadratic equation:

x² - 2.040816 ... x10⁻⁵x - 1.63265 ... x10⁻⁶ = 0

x = 1.287997 ... x10⁻³ = [OH⁻]

The pOH is calculated as follows:

pOH = - log[OH⁻] = - log (1.287997 ... x10⁻³) = 2.890 ...

The pH is calculated from the pOH as follows:

pH = 14 - pOH = 14 - 2.890 = 11.11