Question

How will the pH of solution change when 30 ml of 0.2 M sodium hydroxide solution is added to

300 ml of water​

Answer 1

Answer:

Answer is attached above.

Answer 2

The pH of the solution will change to 12.25

Explanation:

Given:- Sodium Hydroxide = 0.2 M

Now, sodium hydroxide will dissociate into its ions

$NaOH (aq)$  ⇄ $Na^{+} +OH^{-}$

As, millimoles of NaOH

M x V = 30 x 0.2 = 6 m. mol

Therefore, millimoles of $OH^{-}$ formed = 6 m . mol

(as complete dissociation of NaOH takes place)

$[OH^{-}]$ = 6 m. mol     =    6     =   1.8 x $10^{2}$ $molL^{-1}$

            300+ 30ml       330

pOH = - log ( 1.8 x $10^{-2}$ ) = 2 - log ( 1.8) = 2 - 0.25 = 1.75

So, pH = 14 - pOH = 14 - 1.75 = 12.25

SO pH is 12.25