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The enthalpies of neutralization of NaOH...

The enthalpies of neutralization of `NaOH` & `NH_(4)OH` by `HCl` are -13680 Cal and -12270 Cal respectively. What would be the enthalpy change if one gram equivalent of `NaOH` is added to one gram equivalent of `NH_(4)Cl` in solution ? Assume that `NH_(4)OH` and `NaCl` are quantitatively obtained.

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The correct Answer is:
`-1410` Cal
`NaOH + NH_(4)Cl rarr NaCl + NH_(4)OH " Delta H = ?`
`Delta H = - 13680 - (-12270) = - 1410 Cal`
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