Question

what is the pH of 2.5*10^-5 M solution hydroxide pH?

Answer 1

Answer:

12.3

Explanation:

NaOH is  base having  conc 2.5*$10^{-5}$

pOH = -log(OH)

pOH = -log( 2.5*$10^{-5}$)

pOH = 4.6020

pH = 14 - pOH

pH = 14 - 4.6020

pH= 12.3

Answer 2

Given:-

• 2.5*10^-5 M solution hydroxide Solution.

To Find:-

• pH.

Solution:-

Sodium hydroxide - NaOh

We know that NaOh is a strong base

so, it's pH value will more than 7.

$\sf \: NaOh= Na^+ +Oh^-$

So, from this The salt dissociation is 1:1 mole ratio.

As it is in 1:1 ratio

then we can also write it as

$\sf [Oh ^{ - } ] = [NaOh] = 2.5 \times 10^{ - 5} M$

Now, the pH of the solution is determined by the concentration of hydronium ions, $H_3O^-$

we know that ion product constant of water, $\sf \: K_W$

So,

$K_W =[OH ^ −]⋅[H _3O^ +]$

$\: \bf\underline \red{\underline{K_W=10^{-14}}}$

So,

$= > K_W =10^{-14}= [OH ^ −]⋅[H _3O^ +]$

$= > [H _3O^ +] = \frac{10^{-14}}{ [OH ^ −]}$

$= > [H _3O^ +] = \frac{10^{-14}}{ 2.5 \times 10^{ −5}}$

$= > [H _3O^ +] = \frac{10^{-9}}{ 2.5 }$

$= > [H _3O^ +] = \frac{10^{-8}}{ 25 }$

$= > [H _3O^ +] = 0.04 \times 10^{-8} M$

$= > [H _3O^ +] = 4 \times 10^{ - 10} M$

we know that pH of the solution is equal to:

$\: \bf\underline \purple{\underline{pH=−log([H _ 3O ^ +])}}$

$= > pH=−log([4 \times 10^{ - 10} M ])$

$= >pH = 9.39794000867$