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What is the molarity of the solution prepared when #13.50mL# of #8.00molL^-1# #"hydrochloric acid"# is add to #225.0mL# of water?

Answer 1

The new concentration is approx. #0.5*mol*L^-1# with respect to sulfuric acid.

#"Concentration (molarity)"-="Moles of solute"/"Volume of solution"#.

And thus, we have the following.

#(13.50xx10^-3*Lxx8.00*mol*L^-1)/(225.0xx10^-3*L)=0.480*mol*L^-1#.

This value is the formal concentration with respect to #H_2SO_4#; of course we know that sulfuric acid speciates in aqueous solution (to what?).

Conc. acid is always added to water and never the other way around. Why not? Because if you spit in acid, it will spit back at you. I just wanted to say that I agree with your order of addition.

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Answer 2

Peyton Adams

The molarity of the solution is 0.43 mol/L.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

What is the molarity of the solution prepared when #13.50*mL# of #8.00*mol*L^-1# #"hydrochloric acid"# is add to #225.0*mL# of water?

What is the molarity of the solution prepared when #13.50mL# of #8.00molL^-1# #"hydrochloric acid"# is add to #225.0mL# of water?