84.2 ml of 0.120 m HCL are needed to completely neutralize 50.0 ml of 0.101 m Ba(OH)2 solution.
Explanation of the given equation go through the given steps.
- 2HCL + Ba(OH)2= BaCl2 + 2H2O
- Calculating number of moles of Ba(OH)2 in 50 ml of 0.101 m solution.
- This way Ba(OH)2 is neutralized.
- 0.00505 mol Ba(OH)2 x 2 mol HCL/1 mol Ba(OH)2 = 0.0101 mol HCL
- This is how many HCL moles are needed to neutralize the base Ba(OH)2. Since this is the quantity of moles needed to neutralize the base and the volume of solution containing this quantity of moles, we must now determine how many mL the solution contains.
- 0.0101 mol HCL x (1000ml/0.120 mol HCL = 84.2ml of 0.120M HCL
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