200 mL of 1.0 M NaOH at 25 °C is mixed with 150 mL of 1.0 M HCI, also at 25 °C. If the enthalpy of neutralization of HCI with Na OH is 48.8 kJ mol^-1. The temperature of the solution after mixing the two solutions will be (specific heat capacity and density of both the solution are 4.18 J K^-1 g^-1 and 1 g mL^-1 , respectively.)

Amount of Na OH, n₁ = V _M= (0.200 L) (1.0 mol L^-1 ) = 0.200 mol 

Amount of HCI, n₂ = V_M = (0.150 L) (1.0 mol L^-1 ) = 0.150 mol 

Only 0.150 mol of H⁺ will react with OH^-, heat liberated in this process is 

$q=n{\mathrm{\Delta }}_{\text{neut }}{H}^{\circ }=\left(0.150\mathrm{mol}\right)\left(48.8\mathrm{kJ}{\mathrm{mol}}^{-1}\right)=7.32\mathrm{kJ}=7320\mathrm{J}$

 Increase in temperature of the solution will be $\mathrm{\Delta }T=\frac{q}{mc}=\frac{7320\mathrm{J}}{\left(350\mathrm{g}\right)\left(4.18{{\mathrm{Jg}}^{-1}}^{\circ }{\mathrm{C}}^{-1}\right)}={5.0}^{\circ }\mathrm{C}$

Hence, temperature of solution is 25.0 °C + 5.0 °C = 30.0 °C