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.

$(0.001M)\times 250ml=M_2\times (1000ml)$

$M_2=2.5\times 10^{-4}M$

The concentration of $H^+$ ion = $2\times (2.5\times 10^{-4}M)=5\times 10^{-4}M$

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

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

$pH=-\log (5\times 10^{-4})$

$pH=3.30$

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