Question

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

Answer 1

Explanation:

Given pH is 12 so pOH is 2.

Now

$p0h = - log(0h)$

Here we can find

Concentration of OH- ion is 0.01

$mole \: number = molarity \times volume$

thus mole number is (0.01 X 0.5) = 5x10^-3

In 1 mole amount of OH- ion is 6.02x10^23

so in 5x10^-3, there are (6.02x10^23 X 5x10^-3)= 3.01x 10 ^ 21 OH- ions

Answer 2

Answer:

The concentration of the OH⁻ is

Explanation:

Given the pH of the solution is 12.

 volume of the solution is 500ml = 0.5L

Now the pOH of the solution is 14-12 = 2.

We know that

    $pOH = -log([OH^{-} ])$  which is equal to 2

Therefore the concentration of the OH⁻ is 0.01M in 1L

 the concentration of OH⁻ in 0.5L is 0.005M