The enthalpy change when a strong acid is neutralized by strong base is –56.1 kj/mol. if 135 ml of 0.450 m hi at 23.15°c is mixed with 145 ml of 0.500 m naoh, also at 23.15°c, what is the maximum temperature reached by the resulting solution? (assume that there is no heat loss to the container, that the specific heat of the final solution is 4.18 j/g·°c, and that the density of the final solution is that of water.)

First, we have to get the moles of HI = molarity * volume L

                                                              =  0.45 m * 0.135 L

                                                              = 0.06075 moles

then the moles of NaOH = molarity * volume L

                                          = 0.5 m * 0.145 L

                                          = 0.0725 moles

when we have the reaction equation is:

HI + NaOH → H2O + NaI

and HI is the limiting reactant

when Q = molar enthalpy change * moles HI

             = 56100 j / mol * 0.06075 mol 

            = 3408 j

when Q = M * Cp * ΔT

M is the mass & Cp is the specific heat & ΔT is difference in temperature

we need to get the mass (m), when the total volume = 135 ml +145 ml
                                                                                   =280 ml

∴ Mass (m) = volume * density

                   = 280 * 1 = 280 g 

so by substitution:

3408 = 280 g * 4.18 * ΔT

∴ΔT = 2.9 °C

∴(Tf- Ti ) = 2.9 °C

when Ti = 23.15°C 

∴Tf = 23.15 +2.9 

      = 26.05 °C

Rn in the right answer