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

because u need two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei to form isotopes.

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