# Question #8bf92

##### 1 Answer

#### Explanation:

The thing to remember about sulfuric acid is that you can assume that it acts as a **strong acid** in *both ionizations*, not just in the first one.

#"H"_ 2"SO"_ (4(aq)) + 2"H"_ 2"O"_ ((l)) -> color(red)(2)"H"_ 3"O"_ ((aq))^(+) + "SO"_ (4(aq))^(2-)#

This implies that, for all intended purposes, you can say that a solution of sulfuric acid will always contain

#["H"_3"O"^(+)] = color(red)(2) * ["H"_2"SO"_4]#

In your case, the concentration of hydronium cations will be equal to

#["H"_3"O"^(+)] = color(red)(2) * "0.002 M"#

#["H"_3"O"^(+)] = "0.004 M"#

Now, the

#color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))#

This means that your solution will have

#"pH" = - log(0.004)#

# = - log(4 * 10^(-3))#

# = - [log(4)+ log(10^(-3))]#

# = - log(4) - (-3) * overbrace(log(10))^(color(blue)(=1))#

# = 3 - log(4)#

# = color(darkgreen)(ul(color(black)(2.4)))#

The answer is rounded to one *decimal place*, the number of **sig figs** you have for the concentration of the acid.