# Calculate poH of 0.5M Ca(OH)2

## Answers

Answered by

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

3

**Answer:**

**Answer:**

__Given:__

⇒ 0.5 M Ca(OH)₂

__To Find:__

⇒ pOH of 0.5 M Ca(OH)₂

__Solution:__

We know that,

**⇒ Ca(OH)₂ is a Strong Base**

**It Dissociates Completely**

__The Chemical Equation for it's Dissociation is__

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⁻ = **

Concentration of **OH⁻ **= 1

*We know that,*

*Therefore,*

We know that,

*So,*

*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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