Question

The pH of monoacidic weak base is 11.2. Calculate itspercent dissociation in 0.02 M solution.​

Answer 1

$\huge\star\mathfrak\pink{{Answer:-}}$

Solution : pOH of the solution is given as :

pOH = 14 - pH = 14 - 11.2 = 2.8

pOH = -log10[OH ]

log10[OH ] = - pOH

= - 2.8 = - 2 - 0.8 - 1 + 1

= - 3 + 0.2 = 3.2

[OH ] = antilog 3.2 = 1.585 × 10-3 mol/dm3

For monoacidic base,

BOH (aq) B⊕(aq) + OH (aq)

[OH ] = ∝c

∝ = [OH ]c

= 1.585 × 10-3

0.02 = 0.07925

Percent dissociation = ∝ × 100

= 0.07925 × 100

= 7.925 %

Answer 2

Answer:

The percent disassociation is $7.925%$%.

Step-by-step explanation:

We are given that the pH of monoacidic weak base is $11.2$.

We need to determine its percent dissociation in $0.02 M$ solution.​

We know that  pOH of the solution is given by:

$pOH = 14 - pH = 14 - 11.2 = 2.8$

Also, we know that

$pOH = -log10[OH ]$

Solving, we get

$= - 2.8 = - 2 - 0.8 - 1 + 1$

$= - 3 + 0.2$

$= 3.2$

Now,

$[OH ] = antilog 3.2 = 1.585 * 10^{-3} mol/dm^{3}$

For the given monoacidic base,

$[OH ] = \alpha c$

$= 1.585 *10^{-3}$

$0.02 = 0.07925$

So, percent dissociation $= \alpha *100$

$= 0.07925 *100$

$= 7.925 %$

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