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

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

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