Question

Question #d9ade

Answer 1

`"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?