Question

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4. 10 Points. SHOW YOUR WORK CLEARLY a) USE THE FACTOR LABEL METHOD What volume of a 0.500 M HCl solution is needed to neutralize 10.0 mL of a 0.220 M Ba(OH), solution 2 HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + 2 H2O(1) b) USE THE FACTOR LABEL METHOD An experiment calls for the addition of 0.250 g of HCl to reaction vessel. You have at your disposal a 0.250 M HCl solution. How many mL of the 0.250 M HCl should be added? 5. 10 Point SHOW YOUR WORK CLEARLY. a) You have 505 mL of a 0.125 M HCl solution and you want to dilute it to exactly 0.100 M. How much water should you add? b) What mass of Ba(OH)28H20 would be required to make 250.0 mL of a solution that is 1.250 M in hydroxide ions? Use tector lable method.

Answer 1

Write the reaction as follows: 2HCl(aq) + Ba(OH),(aq) →→ BaCl, (aq) + 2H,0(1) Molarity of HCl = 0.500 M volume of HCl = xL moles of HC1 = Molarity of HCl x volume of HCI = 0.500 Mx mol/l -XXL M = 0.500x mol Molarity of Ba(OH), = 0.220 M volume of Ba(OH), = 10.0mL moles of Ba(OH), = Molarity of Ba(OH), x volume of Ba(OH), = 0.220 MX MOHT 10.0 mL x 1000 mL = 0.0022 mol from the above reaction, moles of HCI moles of Ba(OH), Substitute the corresponding values in the above equation and caluclatex 0.500x mol 0.0022 mol 0.0022x2 x= 0.500 1L x = 0.0088 L x1000 mL x = 8.8 mL Therefore, the required solution is 8.8 mL ----------- ----- - - - - - - - - - - - (6) mass of HCl = 0.250 g molar mass of HCl = 36.5 g/mol moles of HC1 = 0.250 gx me = 0.00685 mol 36.5g Molarity of HCl = 0.250 M volume of HCl = x Calculate the volume of HCl as follows: moles of HCl = Molarity of HC] x volume of HCI 0.00685 mol = 0.250 Mxmoll M ** 0.00685 0.250 L x= 0.0274 L x 1000 m x= 27.4mL Therefore, the required solution is 27.4 mL