Respuesta :
Answer : The pH of 0.001 M [tex]H_2SO_4[/tex] 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,
[tex]H_2SO_4\rightarrow 2H^++SO_4^{2-}[/tex]
The concentration of [tex]H_2SO_4[/tex] = x = 0.001 M
The concentration of [tex]H^+[/tex] ion = 2x = 2 × 0.001 M = 0.002 M
The concentration of [tex]SO_4^{2-}[/tex] = x = 0.001 M
Now we have to calculate the pH of 0.001 M [tex]H_2SO_4[/tex].
[tex]pH=-\log [H^+][/tex]
[tex]pH=-\log (0.002)[/tex]
[tex]pH=2.69[/tex]
Now we have to calculate the molarity after mixing the solution.
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of [tex]H_2SO_4[/tex] solution = 0.001 M
[tex]V_1[/tex] = volume of [tex]H_2SO_4[/tex] solution = 250 ml
[tex]M_2[/tex] = molarity of after mixing = ?
[tex]V_2[/tex] = 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.
[tex](0.001M)\times 250ml=M_2\times (1000ml)[/tex]
[tex]M_2=2.5\times 10^{-4}M[/tex]
The concentration of [tex]H^+[/tex] ion = [tex]2\times (2.5\times 10^{-4}M)=5\times 10^{-4}M[/tex]
Now we have to calculate the pH after mixing the solution.
[tex]pH=-\log [H^+][/tex]
[tex]pH=-\log (5\times 10^{-4})[/tex]
[tex]pH=3.30[/tex]
Therefore, the pH of 0.001 M [tex]H_2SO_4[/tex] is, 2.69 and the pH after mixing the solution is, 3.30.
