Question #d9ade

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Answer 1 · 2017-06-18T06:09:09

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

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

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?