Okay so I did this problem, but it seems wrong? I put the answers I got...
Determine the pH of the following tiration at each of the points indicated. A 71.7 mL solution of 0.182 M NaOH is titrated with 0.2086 M HCL.
1.) initial pH= is it 7.86 or 8.51?
2.) after addition of 10.0 mL of HCL= is it 9.98?
3.) after addition of 25.0 mL of HCL= is it 9.50
4.) 50.0 mL of HCL= is it 8.65?
5.) 100.0 mL HCL= is it 5.18?
a and b no.
You have 71.7 x 0.182M NaOH = 13.049 mmoles NaOH.
a. At the initial point, (zero mL HCl) you have 0.182M NaOH. So pOH = -log(OH^-) = -log(0.182) = -0.74 and 14.0=-0.74 = 13.26.
b. You add 10mL x 0.2086M = 2.086 mmoles HCl.
..............NaOH + HCl ==>NaCl + H2O
initial mmoles..13.049..0....0......0
add HCl.............2.086.........
change......-2.086.-2.086..2.086.2.086
equil.......10.963.....0....2.086..2.086
pH is determined by the NaOH since that is in excess. (NaOH) = mmoles/mL = 10.963/(71.7+10)mL = 0.134M
Then pOH = 0.872 and pH = 14.0-0.872 = 13.13.
Etc.
3 is determined by excess NaOH
4 is determined by excess NaOH
5 is determined by excess HCl.
so all of them are wrong XD all right, thanks for the info! I'm going to redo the problem right now
To determine the pH at each point indicated in the titration, we need to understand the concept of acid-base reactions and titration.
1.) The initial pH can be determined by considering the starting solution (before any HCl is added). In this case, we have a 71.7 mL solution of 0.182 M NaOH. NaOH is a strong base and completely dissociates in water, yielding hydroxide ions (OH-) and sodium ions (Na+). Since the solution only contains a strong base, the initial pH will be determined by the hydroxide ion concentration.
To find the hydroxide ion concentration, we can use the equation:
Molarity (M) = moles/volume (L)
Given that the volume is 71.7 mL (0.0717 L) and the molarity is 0.182 M, we can calculate the number of moles of NaOH in the solution:
moles NaOH = 0.182 M x 0.0717 L = 0.01309 mol
Since NaOH completely dissociates, the moles of hydroxide ions (OH-) will be equal to the moles of NaOH. To find the hydroxide ion concentration, we divide the moles by the total volume of the solution:
[OH-] = moles OH-/total volume (L) = 0.01309 mol / 0.0717 L = 0.182 M
To determine the initial pH, we can use the equation:
pOH = -log[OH-]
Since the pOH and pH are related (pOH + pH = 14), we can find the initial pH:
pH = 14 - pOH = 14 + (-log 0.182) = 8.74
Therefore, the correct value for the initial pH is 8.74.
2.) After the addition of 10.0 mL of HCl, we need to calculate the new hydroxide ion concentration to determine the pH. HCl is a strong acid that completely dissociates in water, yielding hydrogen ions (H+) and chloride ions (Cl-).
Using the same approach, we can calculate the moles of HCl in the 10.0 mL (0.01 L) portion.
moles HCl = 0.2086 M x 0.01 L = 0.002086 mol
Since HCl completely dissociates, the moles of hydrogen ions (H+) will be equal to the moles of HCl. Adding the moles of H+ to the moles of OH- obtained from the previous step, we can calculate the new hydroxide ion concentration:
[OH-] = (moles OH-)/(total volume+added volume) = 0.01309 mol / (0.0717 + 0.01) L = 0.153 M
Now we can calculate the new pOH and pH:
pOH = -log[OH-] = -log(0.153) = 0.82
pH = 14 - pOH = 14 - 0.82 = 13.18
Therefore, the correct value after the addition of 10.0 mL of HCl is pH 13.18.
3.) After the addition of 25.0 mL of HCl, you would follow the same procedure as in step 2 to calculate the new hydroxide ion concentration and subsequently the pH.
[OH-] = (moles OH-)/(total volume+added volume) = 0.01309 mol / (0.0717 + 0.01 + 0.025) L = 0.117 M
pOH = -log[OH-] = -log(0.117) = 0.93
pH = 14 - pOH = 14 - 0.93 = 13.07
Therefore, the correct value after adding 25.0 mL of HCl is pH 13.07.
4.) After the addition of 50.0 mL of HCl, you would repeat the same procedure as in step 2 to calculate the new hydroxide ion concentration and determine the pH.
[OH-] = (moles OH-)/(total volume+added volume) = 0.01309 mol / (0.0717 + 0.01 + 0.025 + 0.05) L = 0.080 M
pOH = -log[OH-] = -log(0.080) = 0.91
pH = 14 - pOH = 14 - 0.91 = 13.09
Therefore, the correct value after adding 50.0 mL of HCl is pH 13.09.
5.) Finally, after the addition of 100.0 mL of HCl, repeat the same procedure as in step 2 to calculate the new hydroxide ion concentration.
[OH-] = (moles OH-)/(total volume+added volume) = 0.01309 mol / (0.0717 + 0.01 + 0.025 + 0.05 + 0.1) L = 0.048 M
pOH = -log[OH-] = -log(0.048) = 1.32
pH = 14 - pOH = 14 - 1.32 = 12.68
Therefore, the correct value after adding 100.0 mL of HCl is pH 12.68.
It seems that the initial pH you calculated as 7.86 is incorrect. Additionally, it's important to note that the pH values obtained in a titration experiment are approximate, as actual experimental and calculation errors can occur.