Question

The pOH of a solution of NaOH is 11.30. [H+] in M of this solution is

Answer 1

Answer:

2.0×10^-3M

Explanation:We know,

pH+pOH=14

14-11.30=pH

pH=2.7

Now, pH= -log[H^+]

2.7= -log[H^+]

[H^+] = -antilog(2.7)

[H^+] = 2.0×10^-3

Answer 2

Given: The pOH of a solution of NaOH is 11.30.

To find: [H+] in M of this solution is

Solution: pH tells the amount of H+ ion present in any solution. They are related in such a way that if the concentration of H+ ion is higher then the pH level of that solution will be low that is closer to 0 and hence considered to be more acidic in nature.

To find the pH of any solution with the help of its pOH level we have a formula that is

pH + pOH = 14

on putting values we will get pH

pH = 14-11.30= 2.70

Now we know that pH = - log[ concentration of H+ ]

[H+] = - antilog( 2.7)

[H+] = 2.0×10^-3

Therefore, [H+] in M of this solution is 2×10^-3.