Enthalpy of neutralzation is defined as the enthalpy change when 1 mole of acid `//`base is completely neutralized by base `//`acid in dilute solution .
For Strong acid and strong base neutralization net chemical change is
`H^(+) (aq)+OH^(-)(aq)to H_(2)O(l)`
`Delta_(r)H^(@)=-55.84KJ//mol`
`DeltaH_("ionization")^(@)` of aqueous solution of strong acid and strong base is zero .
when a dilute solution of weak acid or base is neutralized, the enthalpy of neutralization is somewhat less because of the absorption of heat in the ionzation of the because of the absorotion of heat in the ionization of the weak acid or base ,for weak acid /base
`DeltaH_("neutrlzation")^(@)=DeltaH_("ionization")^(@)+ Delta _(r)H^(@)(H^(+)+OH^(-)to H_(2)O)`
under same conditions ,how many mL of 0.1 m NaOH and 0.05 M `H_(2)A` (strong diprotic acid ) solution should be mixed for a total volume of 100mL to producce the hight rise in temperature ?

Options

A. `25 : 75`
B. `50 : 50`
C. `75 : 25`
D. `66.66 : 33.33`

Answer

B

Explanation

`2NaOH+H_(2)ArarrNa_(2)A+2H_(2)O`
`(0.1V_(1))/(1000)" "(0.05V_(2))/(1000)`
`V_(1)+V_(2)=100" "....(1)`
`2xx(0.05 V_(2))/(1000)=(0.1V_(1))/(1000)`
`implies V_(1)=V_(2)`
`" "V_(1)=V_(2)=50mL`