What is the pH of the solution created by combining 1.20 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)? with 8.00 mL of the 0.10 M HC2H3O2(aq)?
mL NaOH pH w/ HCl pH w/ HC2H3O2
1.20 ? ?
Complete the table below:
What are the pH values if you take into account that the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water (like it will be in the experiment you perform in lab)?
mL NaOH pH w/ HCl pH wHC2H3O2
1.20 ? ?
Which point on the titration curve was calculated above?
I worked this twice yesterday and once today. What do you not understand about the problem and I can help you through from there.
where do i start?
You have two problems here. The first is the combination of NaOH and HCl. Work with that one first. The second one is NaOH and HC2H3O2. Work with that one second.
For the first one,
1. Write the balanced equation.
2. Calculate moles NaOH and moles HCl. moles = M x L.
3. Determine which is in excess, if either, and prepare an ICE chart if that will help.
4. pH = -log(H^+).
The second one involves the Henderson-Hasselbalch equation.
To determine the pH of the solution created by combining NaOH and HCl, we first need to understand the concept of a neutralization reaction. When an acid and a base react, they undergo a neutralization reaction that produces water and a salt. The pH of the resulting solution depends on the nature of the salt formed.
Let's start by calculating the number of moles of NaOH and HCl used in the reaction.
Step 1: Calculate the moles of NaOH.
Moles of NaOH = volume (in L) × concentration (in mol/L)
Moles of NaOH = 1.20 mL × (1 L/1000 mL) × 0.10 mol/L
Moles of NaOH = 0.00120 mol
Step 2: Calculate the moles of HCl.
Since HCl is already in the form of a solution, we can directly use the given concentration.
Moles of HCl = volume (in L) × concentration (in mol/L)
Moles of HCl = 8.00 mL × (1 L/1000 mL) × 0.10 mol/L
Moles of HCl = 0.00800 mol
Now that we know the moles of NaOH and HCl used, we can determine the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction and determines the amount of product formed. In this case, the limiting reactant will be the one with the smaller number of moles.
Comparing the moles of NaOH (0.00120 mol) and HCl (0.00800 mol), we see that NaOH is the limiting reactant.
Since NaOH is a strong base and HCl is a strong acid, they react in a 1:1 stoichiometric ratio. So, all the NaOH will react with an equal number of moles of HCl to form water and NaCl. Therefore, we can say that 0.00120 mol of HCl is consumed in the reaction. The excess HCl (0.00800 - 0.00120 = 0.00680 mol) will not react with NaOH.
Now, let's calculate the concentration of HCl remaining in the solution after reacting with NaOH.
Concentration of HCl remaining = (mol of HCl remaining) ÷ (volume of solution)
To obtain the volume of the solution, we need to add the volumes of NaOH and HCl solutions used.
Volume of solution = 1.20 mL + 8.00 mL
Volume of solution = 9.20 mL
Concentration of HCl remaining = 0.00680 mol ÷ (9.20 mL × (1 L/1000 mL))
Concentration of HCl remaining = 0.00680 mol ÷ 0.00920 L
Concentration of HCl remaining = 0.739 mol/L
Now that we know the concentration of HCl remaining, we can calculate the pH of the solution created by combining NaOH and HCl.
The pH of an acid solution can be calculated using the formula:
pH = -log[H+]
Since HCl is a strong acid and fully dissociates in water, its concentration is equal to the concentration of H+ ions. Therefore, the pH of the solution can be obtained by taking the negative logarithm of the H+ ion concentration.
pH = -log[H+]
pH = -log[0.739]
pH ≈ 0.13
So, the pH of the solution created by combining 1.20 mL of 0.10 M NaOH with 8.00 mL of 0.10 M HCl is approximately 0.13.
Now, let's move on to the second part of the question and consider the additional factor of dilution. The given 8.00 mL of 0.10 M acid was diluted with 100 mL of water. This means the concentration of the acid after dilution will change.
To calculate the new concentration, we can use the formula:
C1V1 = C2V2
Where:
C1 = initial concentration
V1 = initial volume
C2 = final concentration
V2 = final volume
Let's substitute the values into the formula:
(0.10 mol/L) × (8.00 mL) = (C2) × (100 mL)
Solving for C2:
C2 = (0.10 mol/L × 8.00 mL) / 100 mL
C2 = 0.008 mol/L
So, after dilution, the new concentration of the acid becomes 0.008 mol/L.
Now, using the same process as before, we can calculate the moles of NaOH and find the limiting reactant:
Moles of NaOH = 1.20 mL × (1 L/1000 mL) × 0.10 mol/L
Moles of NaOH = 0.00120 mol
Comparing the moles of NaOH (0.00120 mol) and the new concentration of HCl (0.008 mol/L), we can see that NaOH is still the limiting reactant.
Therefore, the pH values obtained from the previous calculations (0.13) would remain the same, as they were not affected by the dilution of the acid solution.
Regarding the last part of the question, there is no specific information provided about the point on the titration curve that was calculated above. Additional data such as the volume of the acid or base solution required for complete neutralization is needed to determine the specific point on the titration curve.