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?

Question

asked Jul 11, 2020 52.8k views

1 Answer

Nitish Koundade · Answer 1 · 2020-07-15T09:52:49+0000

Answer : The pH of 0.001 M
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
.

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

$pH=-\log (0.002)$

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
is, 2.69 and the pH after mixing the solution is, 3.30.