Due start of class Wednesday, November 8t SHOW ALL WORK and report answers to th

Question

Due start of class Wednesday, November 8t SHOW ALL WORK and report answers to the appropriate amount of significant figures. 1. The volume of a sample of pure HCl gas is at 189.0 mL at 25.00 and 108.0 mm Hg. This gas was then completely dissolved in 60.00 mL of water and titrated with a barium hydroxide solution. It required 15.70 mL of the barium hydroxide solution to neutralize the HCI. If AHn--118.0 kJ and the temperature of the water and barium hydroxide solution was initially 25.00 °C, what would the final temperature of solution be after titration? Hint: the pressure of the solutions are constant.

Explanation / Answer

Ba(OH)2 + 2HCl = BaCl2 + 2H2O

then

mol of HCl = from ideal gas law

PV = nRT

n = PV/(RT) = (108)(0.189)/(62.4*298) = 0.001097 mol of HCl

assume all HCl reacts

so

Qtotal = mol of HCl * HRxn HCl = 0.001097 * -118 kJ/mol = -0.129446 kJ = 129.44 J

Total mass -> 60 mL of water + 15.70 mL = 75.70 mL; which is approx 75.7 g of water

Q = m*C*(Tf-Ti)

129.44 = 75.7 * 4.184*(Tf-25)

Tf = 129.44 /( 75.7 * 4.184) + 25 = 25.41 °C

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