You have an unlimited amount of 12M HCl stock solution. You need to neutralize 2.00 grams of NaOH by slowly adding 0.100M HCl solution. How will you prepare the proper amount of hydrochloric acid solution to add to the beaker of sodium hydroxide using a 500 ml volumetric flask.
2.00 g = 2.00/molar mass NaOH = about 0.05 moles NaOH and that will take 0.05 moles HCl.
mL12M x M12M = mL0.1M x M0.1M
mL x 12 = 500 x 0.1
mL of 12M acid = (500 x 0.1/12) = 4.167.
I would use a 5.000 mL buret, add exactly 4.167 mL of the 12M stock solution of HCl to the 500 mL volumetric flask, then add DI or distilled water to the mark of the flask and mix thoroughly.
(As extra information, I provide the follwing. For whatever it's worth, I would do it a different way and use a 4 mL pipet, transfer exactly 4.00 mL with a pipet to the flask and proceed. That will NOT make 0.1M; the concn will be 0.096 but that's much simpler. Four mL pipets are easily available (one could use 5 mL pipet for a final concn of 0.12M too) and the cocn is very close to what we wanted. I suppose, however, that the intent of the problem is to make you think how to make an exact 0.1M solution. Another point to remember is that if you intend to titrate the 2.00 g NaOH will a buret, using 0.1M HCl will require L = moles/M = 0.05/0.1 = 0.5 L or 500 mL. Requiring such a large volume using a 50 mL buret means refilling the buret about 5 times and that increases the reading error immensely. We don't like to do that.)
To prepare the proper amount of hydrochloric acid (HCl) solution to add to the beaker of sodium hydroxide (NaOH) using a 500 ml volumetric flask, you need to follow these steps:
Step 1: Determine the balanced chemical equation
Write down the balanced equation for the reaction between HCl and NaOH. In this case, it is:
HCl + NaOH → NaCl + H2O
Step 2: Calculate the amount of HCl needed
Since the equation is balanced in terms of moles, you need to convert the given mass of NaOH into moles. The molar mass of NaOH is 22.99 + 16.00 + 1.008 = 39.02 g/mol.
Using the given mass of NaOH (2.00 grams), divide it by the molar mass of NaOH to find the number of moles:
2.00 g NaOH / 39.02 g/mol NaOH = 0.0512 mol NaOH
From the balanced equation, you can see that the stoichiometric ratio between HCl and NaOH is 1:1. Therefore, you need an equal number of moles of HCl to neutralize the NaOH:
0.0512 mol HCl needed
Step 3: Calculate the volume of 12M HCl solution needed
You are given that the concentration of the stock HCl solution is 12M. The definition of molarity (M) is moles of solute divided by liters of solution. Rearrange this equation to solve for the volume of HCl solution needed:
Volume (L) = moles / Molarity
Volume = 0.0512 mol / 12 M
But this is the volume for the required solution; you need to calculate the volume for the concentrated 12M HCl solution. Let's assume the volume of the concentrated solution needed is V_Concentrated.
0.0512 mol / 12 M = V_Concentrated / 1 L
V_Concentrated = 0.0512 mol * 1 L / 12 M = 0.00427 L
Since the 12M HCl solution is already concentrated, you don't need to dilute or make any further adjustments.
Step 4: Prepare the solution in a 500 ml volumetric flask
Since the calculated volume needed for the concentrated HCl solution is 0.00427 L (4.27 ml), which is less than the capacity of the 500 ml volumetric flask, you can simply add the 4.27 ml of HCl stock solution to the flask using a graduated pipette or syringe. Make sure to transfer it accurately.
Step 5: Add the 0.100M HCl solution
In addition to the concentrated HCl solution, you also need to add a 0.100M HCl solution slowly to neutralize the NaOH. The concentration of this solution is provided. The volume of this solution required is not given, so you need to titrate carefully while monitoring the pH level to ensure complete neutralization.
Remember to always wear appropriate safety equipment, such as gloves and goggles, when handling chemicals in the lab.