Question

What is the pH of 0.001 M H,SO4 (strong acid)? After mixing 250 ml 0.001 M H,SO, with 750 mL water, what is the pH now?

Answer 1

Answer : The pH of 0.001 M
`H_2SO_4` is, 2.69 and the pH after mixing the solution is, 3.30.

Explanation :

First we have to calculate the concentration of hydrogen ion.

The balanced dissociation reaction will be,

`H_2SO_4\rightarrow 2H^++SO_4^(2-)`

The concentration of
`H_2SO_4` = x = 0.001 M

The concentration of
`H^+` ion = 2x = 2 × 0.001 M = 0.002 M

The concentration of
`SO_4^(2-)` = x = 0.001 M

Now we have to calculate the pH of 0.001 M
`H_2SO_4`.

`pH=-\log [H^+]`

`pH=-\log (0.002)`

`pH=2.69`

Now we have to calculate the molarity after mixing the solution.

`M_1V_1=M_2V_2`

where,

`M_1` = molarity of
`H_2SO_4` solution = 0.001 M

`V_1` = volume of
`H_2SO_4` solution = 250 ml

`M_2` = molarity of after mixing = ?

`V_2` = volume of after mixing = 250 + 750 = 1000 ml

Now put all the given values in the above formula, we get the molarity after mixing the solution.

`(0.001M)* 250ml=M_2* (1000ml)`

`M_2=2.5* 10^(-4)M`

The concentration of
`H^+` ion =
`2* (2.5* 10^(-4)M)=5* 10^(-4)M`

Now we have to calculate the pH after mixing the solution.

`pH=-\log [H^+]`

`pH=-\log (5* 10^(-4))`

`pH=3.30`

Therefore, the pH of 0.001 M
`H_2SO_4` is, 2.69 and the pH after mixing the solution is, 3.30.