How many milliliters of 0.130 M HCl are needed to completely neutralize 41.0 mL of 0.107 M Ba(OH)2 solution?
1 answer:
54 mL Ba(OH)2x(0.101 mol Ba(OH)2/1000 mL) x (2 mol OH-/ 1 mol Ba(OH)2 ) = 0.0109 mol OH-
0.0109 mol OH-x (1mol HCl/ 1 mol OH- ) = 0.0109 mol HCl
0.109 mol HCl/(0.130 mol/L HCl) = 0.0839 L HCl
0.0839 L HCl * 1000mL = 83.9 mL of 0.130 M HCl
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