If you have 20.0g of a Maalox tablet, and 40mL of .1 M HCI, how much
extra help from NaOH would you need to bring the acid to neutral? Back titrate using
.05 mL NaOH, drop by drop. Please put this into a formula for me to figure out. Thanks.
To calculate the amount of NaOH needed to neutralize the excess HCl, you will need to use the concept of stoichiometry and the information given.
First, let's determine the number of moles of HCl present in the initial 40 mL of 0.1 M HCl solution:
Number of moles of HCl = volume (in liters) × molarity
= 0.040 L × 0.1 mol/L
= 0.004 moles
According to the balanced chemical equation for the neutralization reaction between HCl and NaOH, 1 mole of HCl reacts with 1 mole of NaOH to form 1 mole of water (H₂O) and 1 mole of NaCl. Therefore, the stoichiometric ratio between HCl and NaOH is 1:1.
Since the volume of NaOH added is given as 0.05 mL and the concentration of NaOH is not provided, we cannot calculate the number of moles of NaOH directly. Hence, we will have to determine it based on the reaction and the stoichiometry.
Assuming the reaction goes to completion:
Number of moles of NaOH = Number of moles of HCl
So, the amount of extra help (NaOH) needed to bring the acid to neutral can be determined as follows:
1. Calculate the number of moles of NaOH from the number of moles of HCl obtained earlier:
Moles of NaOH = Number of moles of HCl = 0.004 moles
2. Use the volume of NaOH added to calculate its concentration:
Concentration of NaOH (in M) = moles of NaOH / volume (in liters)
= 0.004 moles / 0.00005 L
= 80 M
Therefore, if you have added 0.05 mL of a NaOH solution with a concentration of 80 M, drop by drop, you would have used enough NaOH to neutralize the excess HCl.
Please note that the concentration of NaOH might not be realistic or typical for a laboratory setting, as it is extremely concentrated.