Answer
a) $31.1^{\circ}C$
Work Step by Step
Number of moles=
Molarity$\times$Volume of solution in L
$=1.00\,M\times0.050\,L=0.050\,mol$.
0.050 mol $HCl$ and 0.050 mol $NaOH$ are combined.
From table 5.3, we have for one mole $HCl$ and $NaOH$
$\Delta H= -56.2\,kJ/mol$
This implies that in this case,
$\Delta H=0.050\,mol\times -56.2\,kJ/mol$
$=-2.81\,kJ=-2810\,J$
But $\Delta H=q_{p}=-sm\Delta T$
$\implies 2810\,J= 4.184\,J/g\cdot\,^{\circ}C\times100\,g\times\Delta T$
Or $\Delta T= (\frac{2810}{4.184\times100})^{\circ}C=6.716^{\circ}C$
$T_{2}-24.4^{\circ}C=6.716^{\circ}C\implies$
$T_{2}=(24.4+6.716)^{\circ}C=31.1^{\circ}C$
