please explain titration problems. I'm a total noob at this and am trying to answer some prelab questions. Examples:
1. How many mL of a 0.800 M NaOH solution is needed to just neutralize 40 mL of a 0.600 M HCl solution?
2. You wish to determine the molarity of a solution of NaOH. To do this, you titrate a 25.00 mL aliquot of your sample, which has had 3 drops of phenolphthalein indicator so it is pink, with 0.1067 M HCl. The sample turns clear (indication the NaOH has been neutralized with the HCl solution) after the addition of 42.95 mL of the HCl. Calculate the molarity of the unknown concentration of NaOH solution.
Both questions can be answered by substituting the informatin given into:
MaVa = MbVb
Ma = molarity of the acid (HCl)
Va = volume of acid
Mb = molarity of base (NaOH)
Vb = volume of base
The unknown in #1 is Vb
The unknown in #1 is Mb
I want to make a couple of points here.
1. The equation provided by GK works quite well for both of these problems; however, it is good ONLY for those equations in which the mol ratios of acid to base is 1 as they are in HCl and NaOH. If H2SO4 was being titrated with NaOH it would not work (at least without modification).
2. The second point is reserved for the second problem itself, not in the way it was worked. Specifically, I don't like the wording of the problem. I want to point out that the solution was pink initially and it was titrated until the solution was COLORLESS. Colorless is the absence of color; clear is the absence of turbidity. The solution changed COLOR from pink to colorless.
Have a good day.
To understand how to solve titration problems, let's break down the steps involved:
Step 1: Write out the balanced chemical equation:
For example, in problem 1, the balanced equation is:
NaOH + HCl -> NaCl + H2O
In problem 2, the balanced equation is:
NaOH + HCl -> NaCl + H2O
Step 2: Determine the stoichiometry of the reaction:
From the balanced chemical equation, we can see that 1 mole of NaOH reacts with 1 mole of HCl to form 1 mole of NaCl and 1 mole of water.
Step 3: Use the stoichiometry to calculate the moles of the known substance:
In problem 1, we know the volume (in mL) and molarity (in M) of HCl. Multiply the volume by the molarity to calculate the moles of HCl:
moles of HCl = volume of HCl (in L) x molarity of HCl
In problem 2, we know the volume (in mL) and molarity (in M) of HCl. Multiply the volume by the molarity to calculate the moles of HCl:
moles of HCl = volume of HCl (in L) x molarity of HCl
Step 4: Use the stoichiometry to calculate the moles of the unknown substance:
Based on the balanced chemical equation, we know that the ratio of moles of NaOH to moles of HCl is 1:1. Therefore, the moles of NaOH would be equal to the moles of HCl from step 3.
Step 5: Convert the moles of the unknown substance to the desired units:
In problem 1, we want to determine the volume (in mL) of NaOH solution. Using the moles of NaOH from step 4 and the known molarity of NaOH, we can calculate the volume:
volume of NaOH solution (in L) = moles of NaOH / molarity of NaOH
volume of NaOH solution (in mL) = volume of NaOH solution (in L) x 1000
In problem 2, we want to determine the molarity (in M) of NaOH. Using the moles of NaOH from step 4 and the known volume of NaOH solution, we can calculate the molarity:
molarity of NaOH = moles of NaOH / volume of NaOH solution (in L)
Step 6: Substitute the given values into the appropriate equation and calculate the desired value:
In problem 1,
moles of HCl = (40 mL / 1000) L x 0.600 M = 0.024 moles of HCl
moles of NaOH = 0.024 moles of HCl
volume of NaOH solution = 0.024 moles of NaOH / 0.800 M = 0.030 L = 30 mL
In problem 2,
moles of HCl = (42.95 mL / 1000) L x 0.1067 M = 0.00458 moles of HCl
moles of NaOH = 0.00458 moles of HCl
molarity of NaOH = 0.00458 moles of NaOH / (25.00 mL / 1000) L = 0.1832 M
Therefore, the molarity of the unknown concentration of NaOH solution in problem 2 is 0.1832 M.
Sure! Let me explain how to approach titration problems step by step.
1. In a titration problem, you typically have two chemicals of known concentrations that react with each other. The goal is to determine the concentration of one of the solutions by measuring the volume of the other solution required to react completely.
2. The first step is to write the balanced chemical equation for the reaction. In the given example, the balanced equation is:
HCl + NaOH -> NaCl + H2O
3. Next, determine the molar ratio between the reactants from the balanced equation. In this case, the ratio is 1:1, meaning for every 1 mole of HCl, 1 mole of NaOH is required for complete reaction.
4. Now, let's solve the first example.
Given:
Molarity of NaOH solution (M1) = 0.800 M
Volume of HCl solution (V1) = 40 mL = 0.040 L
Molarity of HCl solution (M2) = 0.600 M
Volume of NaOH solution (V2) = ?
Calculate the number of moles of HCl:
Moles of HCl = M2 * V1
Using the molar ratio, we can determine the number of moles of NaOH required:
Moles of NaOH = Moles of HCl
Now, calculate the volume of NaOH solution using its molarity:
V2 = Moles of NaOH / M1
Substitute the values into the equation to get the answer.
5. Let's move on to the second example.
Given:
Volume of NaOH solution (V1) = 25.00 mL = 0.02500 L
Volume of HCl solution (V2) = 42.95 mL = 0.04295 L
Molarity of HCl solution (M2) = 0.1067 M
We need to calculate the molarity of NaOH (M1).
Calculate the number of moles of HCl:
Moles of HCl = M2 * V2
Using the molar ratio, we can determine the number of moles of NaOH:
Moles of NaOH = Moles of HCl
Now, calculate the molarity of NaOH using its volume:
M1 = Moles of NaOH / V1
Substitute the values into the equation to get the answer.
Remember to pay attention to units and to round your final answer to the appropriate number of significant figures.
I hope this explanation helps you tackle your prelab questions. Let me know if you have any further questions or need additional assistance!