A 9.60 mL sample of 12.0 M HCl is used to prepare 125 mL of diluted acid. A sample of 32.4 mL of Ba(OH)2 base is exactly neutralized by 24.0 mL of dilute acid. What is the concentration of the Ba(OH)2 base? Help. I don;t know where to start.
Molarity of diluted acid:
12*(9.6/125)
next, the titration equation:
Na*Va=Nb*Vb
12*(9.6/125)*24=2*Mb*32.4
solve for molarity of base, Mb
Well, well, well! Looks like we've got some chemistry brewing here! Let's break it down.
First things first, we need to find the number of moles of HCl used in the neutralization reaction. To do that, we can use the equation:
moles HCl = Molarity × Volume (in liters)
Since you know the molarity of the HCl solution is 12.0 M and the volume is 9.60 mL (or 0.00960 L), just multiply the two to find the number of moles.
Next up, we can use the balanced chemical equation for the reaction to find the mole ratio between HCl and Ba(OH)2. It looks like this:
2 HCl + Ba(OH)2 ➔ 2 H2O + BaCl2
According to the equation, we need 2 moles of HCl to react with 1 mole of Ba(OH)2. So now we can calculate the number of moles of Ba(OH)2.
Since we have 24.0 mL (or 0.0240 L) of the diluted acid, and we know the molarity of the diluted acid is the same as the original HCl solution, we can use the same formula:
moles Ba(OH)2 = Molarity × Volume (in liters)
However, we need to keep in mind the mole ratio we just discovered. We need twice as many moles of HCl as Ba(OH)2, so we have:
moles Ba(OH)2 = (Molarity × Volume) ÷ 2
Now that we know the number of moles of Ba(OH)2, we can calculate its concentration. Since we have 32.4 mL (or 0.0324 L) of Ba(OH)2, we'll use the formula:
Molarity Ba(OH)2 = moles Ba(OH)2 ÷ Volume (in liters)
Substituting the values we've calculated, we can find the concentration of the base.
Now, I'm impressed that you've made it this far, but don't forget to convert all the volumes to liters and keep track of your units! And voila! You've conquered this chemistry puzzle. Good luck!
To solve this problem, we can use the concept of stoichiometry. Here's a step-by-step solution:
Step 1: Find the number of moles of HCl used to neutralize the Ba(OH)2 solution.
Given:
Volume of HCl solution = 9.60 mL = 0.00960 L (convert to liters)
Molarity of HCl = 12.0 M
Number of moles of HCl = Molarity × volume
= 12.0 M × 0.00960 L
Step 2: Write a balanced chemical equation for the reaction between Ba(OH)2 and HCl:
Ba(OH)2 + 2HCl -> BaCl2 + 2H2O
Step 3: Use the balanced equation to determine the stoichiometry between HCl and Ba(OH)2.
From the balanced equation, we can see that 1 mole of Ba(OH)2 reacts with 2 moles of HCl. Therefore, 2 moles of HCl are required to neutralize 1 mole of Ba(OH)2.
Step 4: Calculate the number of moles of Ba(OH)2 reacted.
Given:
Volume of Ba(OH)2 solution = 32.4 mL = 0.0324 L (convert to liters)
According to the stoichiometry, 2 moles of HCl are required to neutralize 1 mole of Ba(OH)2. Therefore, the number of moles of Ba(OH)2 is half that of HCl:
Number of moles of Ba(OH)2 = (1/2) × number of moles of HCl
Step 5: Calculate the concentration of Ba(OH)2.
Given:
Volume of dilute acid = 24.0 mL = 0.0240 L (convert to liters)
Concentration of Ba(OH)2 = Number of moles of Ba(OH)2 / Volume of Ba(OH)2 solution
Step 6: Substitute the values into the formula:
Number of moles of HCl = 12.0 M × 0.00960 L
Number of moles of Ba(OH)2 = (1/2) × number of moles of HCl
Concentration of Ba(OH)2 = (1/2) × number of moles of HCl / Volume of Ba(OH)2 solution
Step 7: Calculate the concentration of Ba(OH)2.
Now, substitute the values:
Concentration of Ba(OH)2 = (1/2) × (12.0 M × 0.00960 L) / 0.0324 L
Simplify the expression and calculate the solution.
I hope this helps you solve the problem.
To find the concentration of the Ba(OH)2 base, you can use the concept of stoichiometry.
Here are the steps to solve this problem:
Step 1: Find the moles of the diluted acid used in the neutralization.
To do this, we can utilize the given information about the dilute acid. We know that 24.0 mL of the dilute acid is required to neutralize 32.4 mL of Ba(OH)2. Since the volumes of the dilute acid and the base are equal when neutralized, we can assume that the moles of HCl and Ba(OH)2 are also equal.
Using the given concentration of the dilute acid (12.0 M), you can calculate the moles of the diluted acid used in the neutralization:
moles of HCl = concentration of HCl × volume of HCl
moles of HCl = 12.0 M × (24.0 mL / 1000 mL) (converting mL to L)
Step 2: Determine the concentration of the Ba(OH)2 base.
From step 1, we found the moles of HCl used in the neutralization. Since the moles of Ba(OH)2 and HCl are equal, we can use this information to calculate the concentration of the Ba(OH)2 base.
The moles of Ba(OH)2 can be calculated using the mole ratio between Ba(OH)2 and HCl. From the balanced chemical equation for the neutralization reaction:
2 HCl + Ba(OH)2 → 2 H2O + BaCl2
The mole ratio between Ba(OH)2 and HCl is 1:2. Therefore, moles of Ba(OH)2 = 2 × moles of HCl.
Concentration of Ba(OH)2 = moles of Ba(OH)2 / volume of Ba(OH)2
Step 3: Calculate the desired concentration of the Ba(OH)2 base.
In this case, the volume of the Ba(OH)2 base used in the neutralization is given as 32.4 mL. You can now substitute the known values into the equation:
Concentration of Ba(OH)2 = (2 × moles of HCl) / (32.4 mL / 1000 mL) (converting mL to L)
Simplify the equation and calculate the concentration of the Ba(OH)2 base in units of Molarity (M).
Remember to pay attention to significant figures and round your answer to the appropriate number of decimal places.
I hope this explanation helps you understand how to solve the problem. Do let me know if you have any further questions!