HCI + NaOH - Nacl + H 20 HCI = 36.458 NaDH = 39.997 9 200 ML 100 mL tarmom . HCI NaOH 200 ml I m NaOH = 5 NaOH 1 /9 39. 997 9 NaOH 100 ml x I'm HCL = 2.74 HCI m/g 36.445 9 HCI Temperature starts at 19.53ic then NaOH is added and makes the temp go to 21. 66'c . Then HCI is added and it makes the temp go to 23.25 . The heat of this neutralization reaction is AH =-S6 KJ / mol Determine the expected change in temperature. *assume the density of the solution is 1 g/ml and the specific heat capacity is 4 . 184 J / gx c