Question #d9ade

1 Answer
Jun 18, 2017

Answer:

#"Molarity"=0.189*mol*L^-1#. As expected the concentration is almost halved.

Explanation:

#"Molarity"="moles of solute"/"volume of solution"#

And so......

#"moles of solute"="molarity"xx"volume of solution"#.

And if we assume, reasonably, that the volumes are additive, then we get the new concentration by the expression.....

#[H_2SO_4(aq)]=(300.0*mLxx10^-3*L/(mL)xx0.377*mol*L^-1)/(0.600*L)#

#=0.189*mol*L^-1# WITH RESPECT TO #H_2SO_4#.

Now of course we know that sulfuric acid ionizes in aqueous solution, but that is nothing that we need to concern ourselves with here.

What is the #pH# of this solution?