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Answer 1 · 2015-11-24T18:21:27

$0.066"mol/l"$

$H_3PO_(4(aq))+3NaOH_((aq))rarrNa_3PO_(4(aq))+3H_2O_((l))$

No. moles $NaOH=cxxv=0.250xx0.0425=0.01$

From the equation we can see that the no. moles of phosphoric acid must be 1/3 of this:

$:.nH_3PO_4=0.01/3=0.0033$

$c=n/v$

So the concentration of the $H_3PO_4$ is given by:

$c=0.0033/0.05=0.066"mol/l"$