How many mL of 0.3 M HCl(aq) are needed to neutralize a 60 mL solution of 0.15 M Ba(OH)2?
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To determine how many mL of 0.3 M HCl(aq) are needed to neutralize a 60 mL solution of 0.15 M Ba(OH)2, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between HCl and Ba(OH)2.
The balanced chemical equation for the reaction is:
2 HCl(aq) + Ba(OH)2(aq) → 2 H2O(l) + BaCl2(aq)
From the balanced chemical equation, we can see that each mole of Ba(OH)2 reacts with 2 moles of HCl. This means that the molar ratio between HCl and Ba(OH)2 is 2:1.
Step 1: Calculate the number of moles of Ba(OH)2:
Given: Volume of Ba(OH)2 solution = 60 mL = 60/1000 L = 0.06 L
Molarity of Ba(OH)2 = 0.15 M
Number of moles of Ba(OH)2 = Molarity × Volume
Number of moles of Ba(OH)2 = 0.15 mol/L × 0.06 L = 0.009 moles
Step 2: Determine the number of moles of HCl required:
Using the molar ratio from the balanced equation, we know that 1 mole of Ba(OH)2 reacts with 2 moles of HCl.
Number of moles of HCl required = 2 × Number of moles of Ba(OH)2
Number of moles of HCl required = 2 × 0.009 moles = 0.018 moles
Step 3: Calculate the volume of 0.3 M HCl(aq) required:
Given: Molarity of HCl = 0.3 M
Volume of HCl required = Number of moles ÷ Molarity
Volume of HCl required = 0.018 moles ÷ 0.3 mol/L = 0.06 L
Finally, convert the volume to milliliters:
Volume of HCl required = 0.06 L × 1000 mL/L = 60 mL
Therefore, 60 mL of 0.3 M HCl(aq) is needed to neutralize the 60 mL solution of 0.15 M Ba(OH)2.