hey I am really stumped...I would really appreciate some help with this question
1.0560g of antacid is weighed and mixed with 75.00mL of excess 0.1126 M HCL. The excess acid required 4.68mL of 0.1008 M NaOH or back titration Calculate the neutalizing power of the antacid in terms of mmol H+ per gram of antacid
Solve the excess acid first.
MolesExcessAcid= moles NaOH
VolumeExcess*MolariatyEx= .00468L*.1008
solve for volume excess, then subtract it from the original volume, and you can then solve to antacid titration.
To solve this problem, we need to determine the number of moles of HCl and NaOH, and then calculate the neutralizing power in terms of mmol H+ per gram of antacid.
Step 1: Calculate the number of moles of HCl.
Given:
- Volume of HCl solution: 75.00 mL
- Concentration of HCl solution: 0.1126 M
First, convert the volume of HCl solution to liters:
75.00 mL = 75.00 mL x (1 L / 1000 mL) = 0.075 L
Next, calculate the number of moles of HCl using the formula:
moles = concentration x volume
moles HCl = 0.1126 M x 0.075 L = 0.008445 moles
Step 2: Calculate the number of moles of NaOH used in the back titration.
Given:
- Volume of NaOH solution used: 4.68 mL
- Concentration of NaOH solution: 0.1008 M
First, convert the volume of NaOH solution used to liters:
4.68 mL = 4.68 mL x (1 L / 1000 mL) = 0.00468 L
Next, calculate the number of moles of NaOH using the formula:
moles = concentration x volume
moles NaOH = 0.1008 M x 0.00468 L = 0.000471 moles
Step 3: Calculate the number of moles of H+ neutralized by the antacid.
Since NaOH and HCl react in a 1:1 ratio, the number of moles of HCl neutralized by NaOH is equal to the number of moles of H+ neutralized by the antacid.
moles H+ neutralized by the antacid = moles NaOH = 0.000471 moles
Step 4: Calculate the neutralizing power of the antacid in mmol H+ per gram of antacid.
Given:
- Mass of antacid: 1.0560 g
First, convert the mass of antacid to moles using the molar mass of antacid. Let's assume the molar mass of antacid is 100 g/mol (you will need to determine the correct molar mass from your sources):
moles antacid = mass / molar mass
moles antacid = 1.0560 g / 100 g/mol = 0.01056 moles
Finally, calculate the neutralizing power in mmol H+ per gram of antacid:
neutralizing power = (moles H+ neutralized by the antacid / moles antacid) x 1000
neutralizing power = (0.000471 moles / 0.01056 moles) x 1000 = 44.59 mmol H+ per gram of antacid
Therefore, the neutralizing power of the antacid is 44.59 mmol H+ per gram of antacid.
To calculate the neutralizing power of the antacid in terms of mmol H+ per gram of antacid, we need to determine the number of millimoles (mmol) of H+ ions neutralized by the antacid and divide it by the mass of the antacid.
Here's how to get the answer:
Step 1: Determine the number of moles of NaOH that reacted with the excess HCl.
- It is given that 4.68 mL of 0.1008 M NaOH was required.
- Convert mL to L: 4.68 mL = 0.00468 L
- Calculate the number of moles: 0.00468 L * 0.1008 mol/L = 0.000471264 mol NaOH
Step 2: Determine the number of moles of HCl that reacted with NaOH.
- Since the concentration of NaOH is in excess, all the NaOH reacted with HCl.
- The stoichiometry of the reaction is 1:1 (1 mol HCl reacts with 1 mol NaOH)
- Therefore, the number of moles of HCl is also 0.000471264 mol.
Step 3: Determine the number of moles of H+ ions neutralized by the antacid.
- The stoichiometry of the reaction between HCl and antacid is also 1:1. Therefore, the number of moles of H+ ions is 0.000471264 mol.
Step 4: Calculate the mass of the antacid.
- It is given that the mass of the antacid is 1.0560 g.
Step 5: Calculate the neutralizing power of the antacid.
- Divide the number of moles of H+ ions (0.000471264 mol) by the mass of the antacid (1.0560 g) and multiply by 1000 to obtain mmol/g:
(0.000471264 mol H+ / 1.0560 g antacid) * 1000 = 0.446 mmol/g (rounded to three significant figures)
Therefore, the neutralizing power of the antacid is approximately 0.446 mmol H+ per gram of antacid.