Determine the K_b for CN^- at 25 °C.The K_a for HCN is 4.9 × 10^-10. A)4.9 × 10^-14 B)2.3 × 10^-9 C)1.4 × 10^-5 D)2.0 × 10^-5 E)3.7 × 10^-7

Question

Quizplus · Accepted Answer

Using the formula KS1U1B1bS1U1B0 = KS1U1B1aS1U1B0/Keq, we can calculate the Keq for CNS1U1P1-S1S1P0 at 25 °C:

Keq = KS1U1B1aS1U1B0/KS1U1B1bS1U1B0
Keq = (4.9 × 10S1U1P1-10S1S1P0) / KS1U1B1bS1U1B0

Now we need to solve for KS1U1B1bS1U1B0. Rearranging the above equation:

KS1U1B1bS1U1B0 = (4.9 × 10S1U1P1-10S1S1P0) / Keq

We don't have the exact value of Keq, but we can approximate it using the assumption that the reaction is at equilibrium (i.e. the concentrations of products and reactants are constant).

CNS1U1P1-S1S1P0 can be represented by the equation: CN- + S2O3^2- ⇌ CNS1U1P1-S1S1P0

The balanced chemical equation for this reaction shows that the stoichiometry of CN- and S2O3^2- is 1:1. Therefore, at equilibrium, the concentration of CNS1U1P1-S1S1P0 will be equal to the concentration of CN-.

From the problem, we know that the concentration of HCN is 10S1S1P0 M. Since HCN and CN- are conjugate acid-base pairs, we can use the equation Ka x Kb = Kw to find the concentration of CN-.

Ka of HCN = 4.9 × 10S1U1P1-10S1S1P0
Kb of CN- = Kw/Ka = 10S1U1P14-10S1U1P0 / (4.9 × 10S1U1P1-10S1S1P0)
Concentration of CN- = Concentration of HCN = 10S1S1P0 M

Now we can solve for Keq:

Keq = [CNS1U1P1-S1S1P0] / [CN-][S2O3^2-]
Keq = 10S1S1P0 / (10S1S1P0)(10S1S1P0)
Keq = 10S1S1P0 / 10S1U1P0

Plugging in the value for Keq, we get:

KS1U1B1bS1U1B0 = (4.9 × 10S1U1P1-10S1S1P0) / Keq
KS1U1B1bS1U1B0 = (4.9 × 10S1U1P1-10S1S1P0) / (10S1S1P0 / 10S1U1P0)
KS1U1B1bS1U1B0 = 4.9 × 10S1U1P1-10S1S1P0 x 10S1U1P0 / 10S1S1P0
KS1U1B1bS1U1B0 = 4.9 × 10S1U1P1-14S1S1P0

Therefore, the answer is D: 2.0 × 10S1U1P1-5S1S1P0.

Determine the Kb for CN- at 25 °C

Determine the K_b for CN^- at 25 °C