Question

Aqueous sodium hydroxide will react with aqueous hydrochloric acid in an acid-base reaction with a ΔHrxn0 of -55.84 kJ/mole. If 25.0 mL of a 2.15 M hydrochloric solution is mixed with 25.0 mL of a 2.00 M sodium hydroxide solution (both solutions at 23.4 ºC initially), what will be the temperature of the solution when the reaction is over? Assume that the specific heat of all solutions is 4.184 J·g-1·°C-1, and that the density of all solutions is 1.00 g/cm3.

Answer 1

Q: HCL + NaOH - Nacl + Higocus - 55.84 kJ [HC1] = 0.15 M Volume), yel = Q5 mL m. moles of HCl = (Molarity) (rolume) = 0.15 M) ( 25 mb) = 53.75 mimol or 53.75 X10-3 mol n 2e .. [Naoh] = 8 NaOH = 85 ml m.moles of NaOH = $x 25 mL = 50 m. mole or 50x10- 3 mole . ~ - According to the reaction equal moles of Hel and Naoh reacts with each other. Moles of Naon < Moles of Hel Naoh is limiting reagent, will decide the product. Heat released when I mole of her and Naoh reacts = 55.84 kJ Heat released when 50x10-3 moles of Hep and Naoh reacts Completely = (55.84 kb ) (50x10cm.!) = 89.7909 kJ = 8798 Joule

(2) Specific heat of solution is 40ll movemente Mans of solution = Volume * (dentity) blution = (25+45)m! x! som chezim) mi = 50 gm gained by Solution. Heat released by neutralization = Heat 9. = m solution X c solution X Isolution A Trolution Da Moolu ton = 509 Csolution = Heat Copacity = 4.184 I G C AT = TA-T; = To - 093.400 979° J = (50 g) (4.184 ) ( TF-23-4c) Tf - 23.4°C = 13.350 Ti = 36.75 °C →