Question

What is the pH of a solution made by adding 200 mL of distilled water to 100 mL of 0.0030 M HNO3? (assume volume are additive)2.01.03.02.7

Answer 1

The pH of the solution is 3.0

Step-by-step explanation:

Considering:

`Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)`

Or,

`Moles =Molarity * {Volume\ of\ the\ solution}`

Given :

For
`HNO_3` :

Molarity = 0.0030 M

Volume = 100 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 100×10⁻³ L

Thus, moles of
`HNO_3` :

`Moles=0.0030 * {100* 10^(-3)}\ moles`

Moles of
`HNO_3` = 0.0003 moles

Volume added = 200 mL

Total volume = 200 + 100 mL = 300 mL = 0.3 L

Thus,

`Molarity=(0.0003\ moles)/(0.3\ L)=0.001\ M`

SInce, it is strong acid. Thus, [H⁺] = 0.001 M

pH is defined as the negative logarithm of the concentration of hydrogen ions.

Thus,

pH = - log [H⁺] = - log 0.001 = 3.0

The pH of the solution is 3.0