Question

Calculate the [H3O^+] for a solution with pH= 11.50?
A. What is the [H3O+] for a solution with pH = 11.50?
B. What is the [OH^-] for the solution above?
C. What is the [H3O+] for a solution with pH = 5.0?
D. What is the [OH−] for the solution above?
E. What is the [H3O+] for a solution with pH = 6.59?
F. What is the [OH−] for the solution above?
G. What is the [H3O+] for a solution with pH = 1.89?
H. What is the [OH−] for the solution above?

Answer 1

Solving for first part (A):

pH of a solution is the negative base-10 logarithms of hydronium ion concentration.

pH= -log[H3O+]...... (1)

Given, pH= 11.50.

Substituting pH in equation (1).

11.50= - log[H3O+]

[H3O+]= 10^-11.50

= 3.16× 10^-12 M.

We know that, pH+ pOH= 14.

Therefore, pOH= 14- 11.50= 2.50.

pOH= - log[OH-]

2.50= -log[OH-]

[OH-]= 10^-2.50

= 3.162× 10^-3 M.

Solving for second part (B):

pH= 5.0

We have,
pH= -log[H3O+]
Substituting pH= 5.0 in a above equation.
5.0= -log[H3O+]
[H3O+]= 10^-5.0
= 1× 10^-5 M

Now, pH+ pOH= 14.
Therefore, pOH= 14- 5
pOH= 9.
Now,
pOH= -log[OH-]
Substituting pOH value in above mentioned equation, we have
9= -log[OH-]
[OH-]= 10^-9
= 1× 10^-9 M

Calculating for third part (E):

We have,
pH= -log[H3O+]
Substituting pH= 6.59 in a above equation.
6.59= -log[H3O+]
[H3O+]= 10^-6.59
= 2.57× 10^-7 M

Now, pH+ pOH= 14.
Therefore, pOH= 14- 6.59
pOH= 7.41
Now,
pOH= -log[OH-]
Substituting pOH value in above mentioned equation, we have
7.41= -log[OH-]
[OH-]= 10^-7.41
= 3.89× 10^-8 M

Calculating for fourth part (G):

We have,
pH= -log[H3O+]
Substituting pH= 1.89 in a above equation.
1.89= -log[H3O+]
[H3O+]= 10^-1.89
= 1.288× 10^-2 M

Now, pH+ pOH= 14.
Therefore, pOH= 14- 1.89
pOH= 12.11
Now,
pOH= -log[OH-]
Substituting pOH value in above mentioned equation, we have
12.11= -log[OH-]
[OH-]= 10^-12.11
= 7.762× 10^-13 M