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Akimi4 [234]
2 years ago
12

Given that a for HCN is 6. 2×10^−10 at 25 °C. What is the value of b for cn− at 25 °C?

Chemistry
1 answer:
Kay [80]2 years ago
4 0

If Ka for HCN is 6. 2×10^−10 at 25 °C, then the value of Kb for cn− at 25 °C is 1.6 × 10^(-5).

<h3>What is base dissociation constant? </h3><h3 />

The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.

The dissociation reaction of hydrogen cyanide can be given as

HCN --- (H+) + (CN-)

Given,

The value of Ka for HCN is 6.2× 10^(-10)

The correlation between base dissociation constant and acid dissociation constant is

Kw = Ka × Kb

Kw = 10^(-14)

Substituting values of Ka and Kw,

Kb = 10^(-14) /{6.2×10^(-10) }

= 1.6× 10^(-5)

Thus, the value of base dissociation constant at 25°C is 1.6 × 10^(-5).

learn more about base dissociation constant :

brainly.com/question/9234362

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\mathtt{ density \ d  = \dfrac{ 212.81}{47.4540533}}

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