0
Add your answer:
Earn +20 pts
What is KB for cn- aq h2o l hcn aq oh-aq?
The Kb for CN- (aq) is the equilibrium constant for the reaction of CN- with water to form HCN (aq) and OH- (aq). It represents the strength of the base CN- in solution. It can be calculated by taking the concentration of the products (HCN and OH-) and dividing by the concentration of CN- at equilibrium.
An acid has Ka equals 3.1 x 10-10 What is the Kb for the conjugate base of this acid at 25 C?
To find the Kb of the conjugate base, you can use the relationship Kw = Ka * Kb. At 25°C, the value of Kw is 1.0 x 10^-14. Given Ka = 3.1 x 10^-10, you can solve for Kb using Kb = Kw / Ka. This gives you Kb = 1.0 x 10^-14 / 3.1 x 10^-10 = 3.23 x 10^-5.
What is KB for cn aq h2o l hcn aq oh aq?
Kb=[HCN][OH-]/[CN-]
An acid has Ka 3.1 10 10 What is the Kb for the conjugate base of this acid at 25 C Kw 1.0 10 14 at 25 C?
To find the Kb for the conjugate base, you can use the relationship Kw = Ka * Kb. Rearrange the equation to solve for Kb: Kb = Kw / Ka. Plugging in the values, you get Kb = (1.0 x 10^-14) / (3.1 x 10^-10) = 3.23 x 10^-5.
An acid has Ka 2.5 10 4 What is the Kb for the conjugate base of this acid at 25 C Kw 1.0 10 14 at 25 C?
Since Kw = Ka * Kb, we can rearrange the equation as Kb = Kw / Ka. Plugging in the values, Kb = (1.0 x 10^-14) / (2.5 x 10^-4) = 4.0 x 10^-11. This is the Kb for the conjugate base of the given acid.
What is the pka value for NaCN?
NaCN doesn't really have a pKa. In water it becomes Na^+ and CN^-. The CN^- is a base so it will have a Kb and pKb. If you want the pKa of the conjugate acid (HCN), you can find that from 1x10^-14/Kb.
What is KB for cn- aq h2o l hcn aq oh-aq?
The Kb for CN- (aq) is the equilibrium constant for the reaction of CN- with water to form HCN (aq) and OH- (aq). It represents the strength of the base CN- in solution. It can be calculated by taking the concentration of the products (HCN and OH-) and dividing by the concentration of CN- at equilibrium.
An acid has Ka equals 3.1 x 10-10 What is the Kb for the conjugate base of this acid at 25 C?
To find the Kb of the conjugate base, you can use the relationship Kw = Ka * Kb. At 25°C, the value of Kw is 1.0 x 10^-14. Given Ka = 3.1 x 10^-10, you can solve for Kb using Kb = Kw / Ka. This gives you Kb = 1.0 x 10^-14 / 3.1 x 10^-10 = 3.23 x 10^-5.
What is KB for cn aq h2o l hcn aq oh aq?
Kb=[HCN][OH-]/[CN-]
An acid has Ka 3.1 10 10 What is the Kb for the conjugate base of this acid at 25 C Kw 1.0 10 14 at 25 C?
To find the Kb for the conjugate base, you can use the relationship Kw = Ka * Kb. Rearrange the equation to solve for Kb: Kb = Kw / Ka. Plugging in the values, you get Kb = (1.0 x 10^-14) / (3.1 x 10^-10) = 3.23 x 10^-5.
An acid has Ka 2.5 10 4 What is the Kb for the conjugate base of this acid at 25 C Kw 1.0 10 14 at 25 C?
Since Kw = Ka * Kb, we can rearrange the equation as Kb = Kw / Ka. Plugging in the values, Kb = (1.0 x 10^-14) / (2.5 x 10^-4) = 4.0 x 10^-11. This is the Kb for the conjugate base of the given acid.
An acid has Ka equals 2.5 x 10-4 what is the Kb for the conjugate base of this acid at 25C Kw equals 1.0 x 10-14 at 25C?
To find the Kb of the conjugate base, use the relationship Kw = Ka x Kb. Rearrange the equation for Kb: Kb = Kw / Ka = (1.0 x 10^-14) / (2.5 x 10^-4) = 4.0 x 10^-11.
What Kb value represents the weakest base?
Kb = 3.8 10-10
What is the KB value for NAOH?
Kb = 55 It is a very strong base therefore it completely dissociates.
Which of following KB values represent the strongest base?
Kb = 1.8 x 10-5 (apple x)
What is the base-dissociation constant for a weak base?
The base-dissociation constant (Kb) is a measure of the strength of a weak base in water. It indicates the extent to which the base dissociates into its conjugate acid and hydroxide ions in solution. A larger Kb value indicates a stronger base.
An acid has Ka equals 4.5 x 10- 7 What is the Kb for the conjugate base of this acid at 25 C?
2.2 x 10-8
