Question

Calculate poH of 0.5M Ca(OH)2

Answer 1

Answer:

pOH= -log[OH]

since, N= molarity × valency  (m=0.5 , valency(n)=2)

here [OH]= N =  2×0.5 = 1

thus -log[OH]= -log[1] = 0

since Ca(OH)2 is strong base we can confirm the answer

further proving it to be strong base:

* PH= 14- pOH thus pH= 14-0=14 thus it is strong base*

Answer 2

$\bigstar$ Answer:

$\checkmark$ Given:

⇒ 0.5 M Ca(OH)₂

$\checkmark$ To Find:

⇒ pOH of 0.5 M Ca(OH)₂

$\checkmark$ Solution:

We know that,

⇒ Ca(OH)₂ is a Strong Base

It Dissociates Completely

The Chemical Equation for it's Dissociation is

$\blue{\boxed{\boxed{\sf Ca(OH)_2 \longrightarrow Ca^{2+}+2OH^{-}}}}$

From the Chemical Equation,

We can understand that,

1 mol of Ca(OH)₂ gives 2 mol of OH⁻

Given that,

0.5 M Ca(OH)₂

So, the Concentration of OH⁻ is

Concentration of OH⁻ = $\sf 2 \times 0.5$

Concentration of OH⁻ = 1

We know that,

$\red{\boxed{\sf pOH = -log[OH^{-}]}}$

Therefore,

$\implies \sf pOH = -log[1]$

$\purple{\implies \sf pOH = 0}$

We know that,

$\pink{\boxed{\sf pH+pOH=14}}$

So,

$\sf \implies pH=14-0$

$\sf \implies pH=14$

Since,

pOH = 0 and pH = 14

We can Conclude that,

Ca(OH)₂ is a very strong base !!

Hence,

pOH of 0.5 M Ca(OH)₂ is 0

pOH = 0