11. If the pH of a solution increases from 2.0 to 4.0, the H3O^+ ion

concentration

O

a decreases by a factor of 2.

b. decreases by a factor of 100.

c. increases by a factor of 3.

O

d. increases by a factor of 1000,

On increasing the pH of the solution from 2 to 4, the Hydronium ion concentration has been decreased by a factor of 100. Thus, option B is correct.

The log of hydronium ion concentration has been calculated as the pH.

pH = -log [Hydronium ion concentration]

Hydronium ion concentration at pH 2:

2 = -log []

[] = antilog (-2)

[] = 1 M

Hydronium ion concentration at pH 4:

4= -log []

[] = antilog (-4)

[] = 1 M

The relationship in the hydronium ion at pH 2 and pH 4:

pH 2 = 100 pH 4.

Thus on increasing the pH of the solution from 2 to 4, the Hydronium ion concentration has been decreased by a factor of 100. Thus, option B is correct.

For more information about the hydronium ion concentration, refer to the link:

link

Option B. decreases by a factor of 100

Explanation:

To obtain the right answer to the question, let us calculate the concentration of the hydronium ion, [H3O+] in both cases.

For pH = 2,

pH = – Log [H3O+]

2 = – Log [H3O+]

–2 = Log [H3O+]

[H3O+] = antilog (–2)

[H3O+] = 1x10^–2M

For pH = 4,

pH = – Log [H3O+]

4 = – Log [H3O+]

–4 = Log [H3O+]

[H3O+] = antilog (–4)

[H3O+] = 1x10^–4M

Now, we shall determine the ratio of [H3O+] in pH 4 to that of pH 2. This is illustrated below:

[H3O+] in pH 4 : [H3O+] in pH 2 => 1x10^–4 / 1x10^–2 = 0.01 = 1x10^–2

From the above, we can thus say that:

[H3O+] in pH 4 = 1/100 [H3O+] in pH 2

Therefore, the [H3O+] decrease by 100

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

the answer is either b or c