A #20*g# mass of phosphoric acid dissolved in #250.0*mL# aqueous solution, reaches a stoichiometric end-point with WHAT volume of #NaOH(aq)# at #0.120*mol*L^-1# concentration?

1 Answer
Jul 31, 2017

Answer:

We need (i) a stoichiometric equation.....

Explanation:

#H_3PO_4(aq) + 2NaOH rarr Na_2HPO_4 +2H_2O(l)#

Note a stoichiometric endpoint is reached upon the addition of 2 equiv of base.

#"Concentration"# #[H_3PO_4]="Moles of phosphoric acid"/"Volume of solution"#

#=((20.0*g)/(98.08*g*mol^-1))/(250*mLxx10^-3*L*mL^-1)=0.816*mol*L^-1#

And (ii) we need equivalent quantitites of #NaOH# when #20*mL# is tritrated.

#=(0.816*mol*L^-1xx20xx10^-3*L)/(1/2xx0.120*mol*L^-1)=0.272*L#

#=27.2*mL#.

Phosphoric acid is a diacid in aqueous solution.

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