Question

50.0 mL of 0.250 M sodium hydroxide at 19.5 oC was added to 50.0 mL of 0.250M hydrochloric acid also at 19.5 oC in a calorimeter. After mixing the solution temperature rises to 21.21oC. What is the heat of reaction? How do you determine the density of this problem? And how do you solve it?

Answer 1

NaCl HO NaOH HCI Reaction: Neutral ization --> The heat of reaction = Heat of neutralization Data: HCaT, 19.5 °C, V 50 mL, M= 0.250 M MaOH: T, 19.5 °C, V =50 m L, M = 0.250 mw HCI 36.5 g/m ol mw NaOH 40 g/mol Tan21.21 (System absorbs heat released by the neutralization reaction AT 1.71°C 9? AH,? pHCag)? NaOH? We calculate the moles of acid and base: moles of HCI = 0.0125 moles 0.250 mol/L x 0.05L moles of NaOH = 0.250 mol/E x 0.05-L = 0.0125 moles We calculate mass of acid and base: mass of HCI= 0.0125 moles x 36.5 g/m ol= 0.46 g mass of NaOH = 0.50 g 0.0125 moles x 40 g/mol Calculate total mass and total moles: 0.46 g0.50 g = 0.96 g total mass 0.025 g total moles 0.0125 moles + 0.0125 moles Calculate heat of neutralization: q mmHCI (aCeH,O (AT) + mNaOHa Ce H2O (AT) Я, 3D тнCI (a x Ce H20 (AT) + mNaOH (ag) x Ce H2O (AT) q=0.46g x 4.186J/g°C (1.71°C) + 0.5 x 4.186J/g°C (1.71°C) a.= 6.87173 J =-6.87173 J eatreleased bythe neutralization reaction дн, total moles - 6.87173 J AH,=0,025 moles AH= -274.87 J/mol Determine the density mass V 0.46 g p of HCI 50 mL p of HCI 0.0092g/mL 0.50 g p of HCI= 50 mL p of HCI 0.01g/mL