3. Theoretical pH of solutions and deionized water after the addition of strong

Question

3. Theoretical pH of solutions and deionized water after the addition of strong acid and strong base in Part II of the experiment.

Measure 20 mL of buffer solution A using a graduated cylinder and pour the solution into a clean beaker labeled A1. Measure 20 mL of buffer solution B using a clean graduated cylinder and pour the solution into a clean beaker labeled B1. Measure 20 mL of the deionized water sample from part I and pour the solution into a clean beaker labeled W1.

3. Using a clean graduated cylinder, add 10 mL of 0.10 M HCl (aq) to each solution A1, B1 and W1. Mix the solutions well using a glass stir rod making sure to rinse the stir rod in between mixing each solution. Measure and record the pH of all 3 solutions in your notebook.

4. Measure 20 mL of buffer solution A using a graduated cylinder and pour the solution into a clean beaker labeled A2. Measure 20 mL of buffer solution C using a clean graduated cylinder and pour the solution into a clean beaker labeled C2. Measure 20 mL of the deionized water sample from part I and pour the solution into a clean beaker labeled W2.

5. Using a clean graduated cylinder, add 10 mL of 0.10 M NaOH (aq) to each solution A2, C2 and W2. Mix the solutions well using a glass stir rod making sure to rinse the stir rod in between mixing each solution. Measure and record the pH of all 3 solutions in your notebook.

*** only need the second table filled out with theoretical, dont need the first table filled out

Calculated Theoretical Measured Solution Volume of Volume of Calculated pH of pH of Acetic Acid Sodium [Acetic Acid] Sodium (mL) Buffer Buffer Acetate in Buffer Acetate) in (mL) Solution (M) Buffer Solution Solution Solution (M 25 25 B 45 5

Explanation / Answer

I am showing you the method of calculating these answers. I am showing you the calcuations fot the beakers C,B, B1 and C2.

In beaker C, we have 5 mL acetic acid and 45 mL sodium acetate.

Using the Henderson Hasselbach equation:

pH = pKa + log([salt]/[acid])

Now after mixing the acid and salt in the given quantities, salt gets diluted by a factor of 50/45, and acid by a factor of 50/5, so we have:

[salt] = 0.5/(50/45) = 0.45 M

[acid] = 0.5/(50/5 = 0.05 M

pH = 4.74 + log( (0.45)/(0.05) ) = 5.694

For beaker C2, 20 mL of buffer is taken.

So,

moles of salt taken = 0.45*0.020 = 0.009, and

moles of acid taken = 0.05*0.020 = 0.001

moles of NaOH added = molarity*volume = 0.1*0.01 = 0.001 moles

Adding these many moles increases the moles of salt by this amount and decreases the moles of acid by this same amount.

So finally we have 0.01 moles of salt and no acid is left.

Final volume is also 30 mL now.

So, pH is established by the following equilibrium:

CH3COO- + H2O <----> CH3COOH + OH-

Initial (0.01/0.03) 0 0

Eqb (0.01/0.03) - x x x

For this reaction,

Kb = ([CH3COOH]*[OH-])/[CH3COO-]

Also, Ka*Kb = 10-14, and Ka = 1.82*10-5

Using all these values and solving we get:

[OH-] = x = 1.352*10-5 M

So,

pOH = 5 - log(1.352) = 4.869

Thus, pH = 14 - pOH = 9.131

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