A 5.00 L solution has [HC7H5O2] = 0.100 M and [Ca(C7H5O2)2] = 0.200 M . What is the pH of the solution after 10.0 mL OF 5.00 M NaOH is added? Assume the volume of the solution does not change .Ka for HC7H5O2 = 6.3x10^-5.
Let's simplify this. HC7H5O2 is benzoic acid = HBz = 0.1 M
Ca(C7H5O2)2 is calcium benzoate = Ca(Bz)2 = 0.2 M
You have mols HBz = M x L = 0.1 x 5 = 0.5 mol = acid
and Bz^- from Ca(Bz)2 = 0.4 x 5 = 2 mols = base
After adding 10 mL of 5 M NaOH = 0.01 x 5 = 0.05 mols then
.................HBz + OH^- ==> Bz^- + H2O
I................0.5..........0...........2.....................
add........................0.05.........................................
C............-0.05......-0.05......+0,05
E.............0.45...........0.......2.05
Plug the E values into the Henderson-Hasselbalch equation and solve for pH.
Post your work if you get stuck.
Well, well, well, looks like we've got a chemical party going on here! Let's see if I can add some comedic flavor to this equation.
First things first, we need to figure out what's happening in this solution. We've got an acetic acid, HC7H5O2, and its calcium salt, Ca(C7H5O2)2, hanging out in some water. And suddenly, NaOH bursts through the door, ready to join the fun!
Now, we know that when NaOH, a strong base, meets up with HC7H5O2, a weak acid, it's going to have a bit of a tango. The NaOH will donate its hydroxide ions (OH-) to the weak acid, forming water and the salt Ca(C7H5O2)2.
But, before we get into the nitty-gritty, let's make a quick calculation to determine how many moles of NaOH we've got. We've got 10.0 mL of a 5.00 M NaOH solution, so the number of moles can be found by multiplying the concentration (5.00 M) by the volume (10.0 mL). Convert that to liters, and we've got ourselves the number of moles of NaOH. I'll let you do the math.
Alright, now it's time to determine the new concentration of HC7H5O2 and Ca(C7H5O2)2 in the solution. The amount of HC7H5O2 that reacted with the NaOH will also be neutralized, leaving us with a new concentration. So, subtract the number of moles of NaOH from the original concentration of HC7H5O2 and Ca(C7H5O2)2.
We're almost there, my friend! Now, we just need to calculate the concentration of OH- in the solution, and then we can figure out the pH. Since NaOH is a strong base, it completely dissolves in water to form OH- ions. So, add the number of moles of NaOH to the volume of the solution, and we'll get the new OH- concentration.
Lastly, we use the equation for acid dissociation constant (Ka) to calculate the concentration of H3O+ ions in the solution. That equation is: Ka = [H3O+][A-]/[HA]. Plug in the known values, and solve for [H3O+]. Once you've got that, you can calculate the pH using the equation: pH = -log[H3O+].
And there you have it, my friend! You've successfully calculated the pH of the solution after adding NaOH. Now, go forth and spread your chemical knowledge with a side of humor!
To determine the pH of the solution after adding 10.0 mL of 5.00 M NaOH, we need to consider the acid-base reaction between NaOH and HC7H5O2.
The balanced equation for the reaction is:
NaOH + HC7H5O2 -> NaC7H5O2 + H2O
First, calculate the moles of HC7H5O2 before adding NaOH:
moles of HC7H5O2 = [HC7H5O2] × volume of solution
moles of HC7H5O2 = 0.100 M × 5.00 L
moles of HC7H5O2 = 0.500 moles
Next, calculate the moles of NaOH added:
moles of NaOH = [NaOH] × volume of NaOH added
moles of NaOH = 5.00 M × 0.010 L
moles of NaOH = 0.050 moles
Since NaOH and HC7H5O2 react in a 1:1 ratio, the moles of HC7H5O2 remaining after the reaction will be:
moles of HC7H5O2 remaining = moles of HC7H5O2 before - moles of NaOH added
moles of HC7H5O2 remaining = 0.500 moles - 0.050 moles
moles of HC7H5O2 remaining = 0.450 moles
Now, calculate the concentration of HC7H5O2 after the reaction:
concentration of HC7H5O2 = moles of HC7H5O2 remaining / volume of solution
concentration of HC7H5O2 = 0.450 moles / 5.00 L
concentration of HC7H5O2 = 0.090 M
Since NaOH is a strong base, it completely reacts with HC7H5O2 and forms the conjugate base C7H5O2-. From the balanced equation, we can see that 1 mol of NaOH produces 1 mol of the conjugate base C7H5O2-. Therefore, the concentration of C7H5O2- after the reaction will also be 0.050 M.
To determine the pH of the solution, we need to calculate the concentration of H+ ions. Since HC7H5O2 is a weak acid, we can use the given Ka value to calculate the concentration of H+ ions:
Ka = [H+][C7H5O2-] / [HC7H5O2]
[H+] = (Ka × [HC7H5O2]) / [C7H5O2-]
[H+] = (6.3 × 10^-5) × (0.090 M) / (0.050 M)
[H+] = 1.134 × 10^-4 M
Finally, we can calculate the pH using the formula:
pH = -log[H+]
pH = -log(1.134 × 10^-4)
pH ≈ 3.95
Therefore, the pH of the solution after adding 10.0 mL of 5.00 M NaOH is approximately 3.95.
To find the pH of the solution, we need to consider the reaction that occurs when NaOH is added to the solution. NaOH is a strong base and will react with the weak acid HC7H5O2. The reaction can be represented as follows:
HC7H5O2 + NaOH → NaC7H5O2 + H2O
The reaction equation tells us that for every one mole of HC7H5O2 that reacts, one mole of NaC7H5O2 and one mole of water are formed.
Since we are given the initial concentrations of [HC7H5O2] and [Ca(C7H5O2)2], we can calculate the initial number of moles of HC7H5O2 and Ca(C7H5O2)2 present in the solution.
moles of HC7H5O2 = concentration (M) x volume (L) = 0.100 M x 5.00 L = 0.500 moles
moles of Ca(C7H5O2)2 = concentration (M) x volume (L) = 0.200 M x 5.00 L = 1.000 moles
Since the reaction between HC7H5O2 and NaOH is a 1:1 reaction, the moles of HC7H5O2 consumed will be equal to the moles of NaOH added. Therefore, we need to calculate the moles of NaOH added.
moles of NaOH added = concentration (M) x volume (L) = 5.00 M x 0.010 L (10.0 mL = 0.010 L) = 0.050 moles
Now, let's determine the limiting reactant. The limiting reactant is the one that will be consumed completely, and the remaining reactant will be in excess. In this case, the limiting reactant will be the one with the fewest moles, which is HC7H5O2.
Since HC7H5O2 and NaOH react in a 1:1 ratio, the number of moles of HC7H5O2 consumed will be equal to the number of moles of NaOH added, i.e., 0.050 moles.
After the reaction, we will have:
moles of HC7H5O2 remaining = initial moles - moles consumed = 0.500 moles - 0.050 moles = 0.450 moles
moles of Ca(C7H5O2)2 remaining = 1.000 moles (unchanged)
Now, we need to calculate the new concentrations of HC7H5O2 and Ca(C7H5O2)2 in the solution.
new [HC7H5O2] = moles remaining / volume = 0.450 moles / 5.00 L = 0.090 M
new [Ca(C7H5O2)2] = moles remaining / volume = 1.000 moles / 5.00 L = 0.200 M (unchanged)
We can now calculate the concentration of OH- ions in the solution, which is equal to the concentration of NaOH added.
[OH-] = concentration (M) = 5.00 M
To calculate the concentration of H+ ions, we'll use the equilibrium expression for the dissociation of HC7H5O2:
Ka = [H+][C7H5O2-] / [HC7H5O2]
We know that the concentration of [HC7H5O2] is 0.090 M (after the addition of NaOH). We can assume that the concentration of [C7H5O2-] is equal to the concentration of [Ca(C7H5O2)2], which is 0.200 M.
Plugging in these values, we have:
6.3x10^-5 = [H+][0.200] / 0.090
Rearranging the equation and solving for [H+], we get:
[H+] = (6.3x10^-5)(0.090) / 0.200
[H+] = 2.835x10^-5 M
Finally, to find the pH of the solution, we can use the equation:
pH = -log[H+]
pH = -log(2.835x10^-5)
pH ≈ 4.55
Therefore, the pH of the solution after adding 10.0 mL of 5.00 M NaOH is approximately 4.55.