What is the concentration of #85%# #H_3PO_4# in #mol*L^-1#?

1 Answer
Apr 15, 2017

Answer:

You need to quote the density of the phosphoric acid.............and you need to add the conc. acid to water...........

Explanation:

This site quotes #rho=1.685*g*mL^-1# for #85%# #H_3PO_4(aq)#.

And thus #"mass of phosphoric acid"# #=# #85%xx1.685*g*mL^-1#, with respect to a #1*mL# volume.........

And we use these data to work out the concentration in #mol*L^-1#.

#rho="Mass of phosphoric acid"/"Volume of solution"#

And thus #"mass of phosphoric acid"=rhoxx"volume of solution"#

#((85%xx1.685*g*mL^-1xx1*mL)/(97.99*g*mol^-1))/(1xx10^-3*L)=14.6*mol*L^-1#.

Given this, we can now work out the required volumes.........

#C_1V_1=C_2V_2#, where #C_2=6.0*mol*L^-1, V_2=100*mL, V_1=??#

#V_1=(C_2V_2)/C_1=(6.0*mol*L^-1xx100*mL)/(14.6*mol*L^-1)=41.1*mL#.

So put approx. #50*mL# of water in a flask, add the required volume of acid, and then make up to #100*mL# volume. You find the EXACT concentration of your solution by titration with standard base, knowing that you have a solution of #Na_2HPO_4(aq)# at the end point.