Question #27dc1

1 Answer
May 22, 2017

Answer:

You add the acid to the water, and never the reverse..........

Explanation:

We want a #1*L# volume of #HCl# solution in water (of course I need to make these assumptions in order to answer the question).

By definition #pH=-log_(10)[H_3O^+]#

And thus #[H_3O^+]=10^-3*mol*L^-1#

And so I need to take a #1*mL# volume of #HCl#, whose concentration is #1.0*mol*L^-1#, and add this a #1*L# volumetric flask full of #0.9*L# water, and then make the volume to #1*L#.

#"Concentration"="Moles of solute"/"Volume of solution"#

#=(1.0xx10^-3*Lxx1*mol*L^-1)/(1*L)=10^-3*mol*L^-1#.

#pH=-log_10(10^-3)=-(-3)=+3#, as required..................