How many mL of 0.655 M phosphoric acid solution are required to neutralize 15 mL of 1.284 M #NaOH# solution?

2 Answers
anor277
Jan 4, 2018

Well, you need to know that phosphoric acid is a DIACID in aqueous solution... The third proton is rarely removed.

Explanation:

And we write the stoichiometric equation as...

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

And so .................

#n_(NaOH)=15*mLxx10^-3*L*mL^-1xx1.284*mol*L^-1=0.01926*mol#...and so we need HALF of this molar quantity, i.e. #9.63xx10^-3*mol#, with respect to #H_3PO_4#.

And so we take the quotient....#(9.63xx10^-3*mol)/(0.655*mol*L^-1)=14.7xx10^-3*L=14.7*mL# to get the volume required for equivalence.

Ratnesh Bhosale
Jan 4, 2018

#V=9.80 mL#

Explanation:

Moles of #NaOH# #= 1.284/1000 xx 15#

Moles of #NaOH# #= 0.01926#

Reaction will be -
#H_3PO_4 + 3NaOH rarr Na_3PO_4 +3H_2O #

So, #0.01926 /3# moles are required to neutralize #H_3PO_4#

Moles required to neutralize #H_3PO_4 = 0.00642#

Let the volume required to neutralize #H_3PO_4# be #V# mL.

Moles of #H_3PO_4# #= 0.655/1000xxV#

#0.00642 = 0.655/1000xxV#

#V= 0.00642xx 1000/0.655#

#V= 6.42/0.655#

#V=9.80 mL#

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