How many milliliters of 0.120 M HCl are needed to completely neutralize 60.0 mL of 0.105 M Ba(OH)2 solution?
Molarity = No.of moles / Volume in litre
Hence, for Ba(OH)2, no. of moles = Molarity X Volume in litre
no. of moles = 0.105 X (60/1000)
= 6.3x10-3 moles
Now, Using Law of Equivalent M1V1 = M2V2
Here, let M1V1 is for Ba(OH)2 and M2V2 for HCl, then this equation becomes: M1V1 = 2 M2V2 as for each molecule of Ba(OH)2, two molecule of HCl are consumed to neutralize it.
Hence, volume of HCl required = M1V1 / 2 M2
= 6.3x10-3 / ( 2x0.120M)
= 0.02625 L
= 26.25 mL
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