We conducted a lab with antacids in class the other day and i'm having trouble doing the calculations. Well we basically took antacid tablets dissolved them in HCl, heated it and then added phenolthalien and then titrated until we saw a pink color. At 41mL we saw the pink and stopped. We are supposed to find the following:
1) Initial HCl moles
2) Moles of HCl neutralized by NaOH
3) Moles of HCl neutralized by antacid
4) Acid neutralizing capacity
We have the following information
1)0.10M NaOH
2)0.10M HCl
3)Mass of Antacid .22g
4)Volume of NaOH titrated from burette 41mL
For the inital moles I got .005 mol HCl, but I get stuck onwards. I know how to get the last one because I have a formula but how do I find the rest of the information? Someone please help!
1. moles HCl initially = MHCl x LHCl
2. moles NaOH added = MNaOH x LNaOH
3. 1-2 = 3
To calculate the required values in your lab experiment, let's break it down step by step:
1) Initial HCl moles:
Since you have 0.10M HCl and a volume of 41mL, you can calculate the number of moles of HCl using the formula:
number of moles = concentration (M) × volume (L)
Converting the volume to liters, we get:
41 mL = (41/1000) L = 0.041 L
Using the formula:
moles of HCl = 0.10 M × 0.041 L
= 0.0041 mol
So, the initial moles of HCl is 0.0041 mol.
2) Moles of HCl neutralized by NaOH:
To find the moles of HCl neutralized by NaOH, we need to use the concept of stoichiometry. The balanced chemical equation between HCl and NaOH is:
HCl + NaOH -> NaCl + H2O
According to the equation, for every 1 mole of HCl, 1 mole of NaOH is required to neutralize it. Since you have the volume (41 mL) of NaOH used, you can calculate the moles of NaOH using the formula mentioned earlier:
moles of NaOH = 0.10 M × (41/1000) L
= 0.0041 mol
Therefore, the moles of HCl neutralized by NaOH is also 0.0041 mol.
3) Moles of HCl neutralized by antacid:
To find the moles of HCl neutralized by the antacid, you need to subtract the moles of NaOH used from the initial moles of HCl:
moles of HCl neutralized = Initial moles of HCl - Moles of NaOH
= 0.0041 mol - 0.0041 mol
= 0 mol
Since the moles of NaOH used is equal to the initial moles of HCl in this case, there are no moles of HCl neutralized by the antacid.
4) Acid neutralizing capacity:
The acid neutralizing capacity is a measure of how well an antacid neutralizes acid. It is the equivalent number of moles of HCl neutralized by the antacid. In this case, the antacid did not neutralize any HCl, so the acid neutralizing capacity is zero.
To summarize, the calculated values are:
1) Initial HCl moles: 0.0041 mol.
2) Moles of HCl neutralized by NaOH: 0.0041 mol.
3) Moles of HCl neutralized by antacid: 0 mol.
4) Acid neutralizing capacity: 0.
Remember, these calculations are specific to your experimental setup, concentrations, and volumes used. Make sure to adjust them accordingly if your experimental conditions are different.