Okay so i have this titration question and I don't really know the steps because everyone explains it differently. IF someone could explain me the steps of how i could solve this questions and other like this. Thanks
In three runs of titration 22.8mL, 221.mL, 22.2mL of 0.200 Ba(OH)2 were required to neutralize 10mL of HCl. Calculate the acid concentration
Step 1. Write and balance the equation.
Ba(OH)2 + 2HCl ==> BaCl2 + 2H2O
Step 2.
mols Ba(OH)2 = M x L = ? That's approx 0.0046 but you need an accurate answer.
Step 3.
Using the coefficients in the balanced equation, convert mols Ba(OH)2 to mols HCl. That's
approx 0.0046 mols Ba(OH)2 x (2 mols HCl/1 mol Ba(OH)2) = approx 0.0092.
Step 4.
M HCl = mols HCl/L HCl = approx 0.0092/0.01 = approx 0.9 M for HCl.
Notes:
1. You can make it easier by working in millimoles. In order of above, 22.8 x 0.2 = approx 4.6.
4.6 x 2 = 9.2. Then M = mollimoles/mL = 9.2/10 = about 0.92M
2. In practice, you can work this three times (by the way I think that 221 mL must be 22.1 mL), then average the three values you get for the final molarity OR, and this easier, average the three mL [(22.8+22.1+22.2/3)] and use that avg mL.
Post your work if you have questions on this. These four steps will work any titration problem like this.
Thank you so much! this helped a lot!
To solve this titration question and similar ones, you can follow these steps:
Step 1: Write the balanced chemical equation.
First, write the balanced chemical equation for the reaction between Ba(OH)2 and HCl. The balanced equation is:
Ba(OH)2 + 2HCl -> BaCl2 + 2H2O
Step 2: Identify the stoichiometry of the reaction.
The balanced equation tells us the molar ratio between Ba(OH)2 and HCl is 1:2. This means that for every 1 mole of Ba(OH)2, we need 2 moles of HCl to react completely.
Step 3: Calculate the moles of HCl used in each run.
To calculate the moles of HCl used in each run, we need to use the formula:
moles = volume (in liters) x concentration (in mol/L)
For the given volumes of HCl in each run (10 mL), the concentration is unknown. Let's call the concentration of HCl "x".
For the first run, the moles of HCl used can be calculated as:
moles (HCl) = volume (in liters) x concentration (in mol/L)
= 0.010 L x x (mol/L)
= 0.010x mol
For the second run, the moles of HCl used can be calculated the same way:
moles (HCl) = 0.010 L x x (mol/L)
= 0.010x mol
And for the third run:
moles (HCl) = 0.010 L x x (mol/L)
= 0.010x mol
Step 4: Calculate the moles of Ba(OH)2 used in each run.
Since the molar ratio between HCl and Ba(OH)2 is 2:1, we can use the moles of HCl used to calculate the moles of Ba(OH)2 used.
For the first run, the moles of Ba(OH)2 used can be calculated as:
moles (Ba(OH)2) = 0.010x mol / 2
= 0.005x mol
For the second run:
moles (Ba(OH)2) = 0.010x mol / 2
= 0.005x mol
And for the third run:
moles (Ba(OH)2) = 0.010x mol / 2
= 0.005x mol
Step 5: Calculate the average moles of Ba(OH)2 used.
To calculate the average moles of Ba(OH)2 used, add up the moles of Ba(OH)2 from each run and divide by the number of runs (3 in this case).
Average moles (Ba(OH)2) = (0.005x + 0.005x + 0.005x) / 3
= (0.015x) / 3
= 0.005x mol
Step 6: Calculate the acid concentration.
Since 0.005x moles of Ba(OH)2 are required to neutralize 10 mL of HCl, we can set up the following equation:
0.005x mol = 10 mL x concentration (in mol/L)
Simplifying the equation:
0.005x = 10 x concentration
concentration = 0.005x / 10
concentration = 0.0005x mol/L
Thus, the acid concentration is 0.0005x mol/L.
Note: In order to solve for x and obtain the specific acid concentration value, you would need additional information or data from the problem statement.