Question

the amount of oh- ion present in 500ml solution of ph = 12​

Answer 1

Answer:

pOH=14−pH

=14−12

=2

pOH=−log[OH

−

]

=2

[OH]

−

=0.01 M

1 litre of solution contains 0.01 mole of NaOH.

0.5 litre of solution contains 0.005 mole of NaOH.

The molecular weight of NaOH is 40 g.

Thus 0.05 mole corresponds to 0.2 g.

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Answer 2

Answer:

At $pH = 12$, the amount of $OH^{-}$ ions present in 500ml of the solution is $0.5\times 10^{-2} M$.

Explanation:

Given,

The volume of the solution = $500ml$ = $0.5L$

The pH of the solution = $12$

The amount of $OH^{-}$ ions present in the solution =?

As we know,

• $pH + p0H = 14$   -------equation (1)

Here,

• pH is known as the power of hydrogen and determines the concentration of free hydrogen ions present in the solution.
• pOH determines the concentration of free hydroxyl ions present in the solution.

After putting the given values in the equation (1), we get:

• $12 + p0H = 14$
• $p0H = 2$

Also,

• $pOH = - log [OH^{-} ]$
• $[OH^{-} ] = 10^{-pOH}$

Now, after putting the value of $p0H = 2$ in the equation, we get:

• $[OH^{-} ] = 10^{-2}$

Therefore,

• The amount of $OH^{-}$ present in 1 L solution = $10^{-2}$M

Hence,

• The amount of $OH^{-}$ present in 0.5L solution = $0.5\times 10^{-2} M$.