Chemistry, asked by nazalways2541, 9 days ago

Calculate poH of 0.5M Ca(OH)2

Answers

Answered by ABI0
0

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*

Answered by BrainlyRonaldo
3

\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

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