Question

(D) TU. 1
4. If 0.02 M Methylamine gets dissociated
to an extent of 15.8%, calculate pOH​

Answer 1

Answer:

sorry I don't read chemistry so sorry

I hope can you understand

Answer 2

Given,

Molarity of Methylamine = 0.02 M

Percentage of dissociation = 15.8%

To find,

pOH of Methylamine

Solution,

• Methylamine is a base which is $NH_{4}OH$.
• Now, $NH_{4} OH$ dissociates into $NH_{4}$ and $OH$.
• It is given that Methylamine gets dissociated to an extent of 15.8% and its molarity is 0.02 M.
• Here 15.8% is dissociated so, 0.02 x $\frac{15.8}{100}$ = 0.00316.
• So $NH_{4}OH$ is dissociated at 0.02 - 0.00316 = 0.01684 M.
• $NH_{4}$ and $OH$ are dissociated at 0.00316 M.
• Since $OH^{-}$ concentration is 0.00316 M = 316 x $10^{-5}$  
• As , pOH =  - log [$OH^{-}$]
• pOH = -log[316 x $10^{-5}$]
• pOH = -log(316) -log($10^{-5}$)
• pOH = -2.499-(-5)
• pOH = 2.501

Therefore, the pOH of 0.02 Methylamine which gets dissociated to an extent of 15.8% is 2.501.