Question

1.How does adding NaOH to deionized water affect the pH differently than adding it to a buffer?

2. how would you find the Ka of solution 4, after it is diluted from solution 3? the measured pH was 4.98 which gave a [H+] concentration of 1.05x10^-5.

Part A - Determining the Acid Ionization Constant for Acetic Acid Solution 1 6. Take a 30.0 mL sample of 10 M HOAc and place it in a beaker. Using a pH meter, measure the pH of the solution and record it in your lab notebook. 7. Add three drops of Universal Indicator to the solution. Using the color chart provided, estimate the pH of the solution. Record the color and estimated pH in your lab notebook. Du Solution 2 8. Take a 30.0 mL sample of 0.1 M HOAc and place it in a beaker. Using a pH meter, measure the pH of the solution and record it in your lab notebook. 9. Add three drops of Universal Indicator to the solution. Record the color and estimated pH in your lab notebook Solution 3 10. Add 24.0 mL of 1.0 M Na Ac to 10.0 mL of 1.0 M HOAc in a beaker and mix well. Using a pH meter, measure the pH of the solution and record it in your lab notebook. 11. Add three drops of Universal Indicator to the solution. Record the color and estimated pH in your lab notebook. Solution 4 12. Dilute 10.0 mL of Solution 3 with 40.0 mL of deionized water in another beaker and mix thoroughly. Using a pH meter, measure the pH of the solution and record it in your lab notebook. 13. Add three drops of Universal Indicator to the solution. Record the color and estimated pH in your lab notebook. Solution 5 14. Take a 30.0 mL sample of 1.0 M NaOAc and place it in a beaker. Using a pH meter, measure the pH of the solution and record it in your lab notebook. 15. Add three drops of Universal Indicator to the solution. Record the color and estimated pH in your lab notebook. Data Analysis For each of the five solutions, calculate the equilibrium [H] from the measured pH and determine the K, value. Two important things to keep in mind are that NaOAc is completely dissociated in aqueous solution, and that dilution occurs when two volumes are mixed, which must be taken into account when calculating the initial concentrations. Additionally, since this Solution 5 did not initially contain any HOAc, Eq. 16-2 cannot be used to calculate Ks; it must be found by first determining Ko from Eq. 16-6. Report the average K, for the solutions and determine the percent error of this average from the accepted value given in the introduction.

Answer 1

1.

Adding NaOH to deionised water means that NaOH, being a strong base, will dissociate completely into Na+ and OH- ions. The OH- ions thus formed will raise the pH according to the equations:

pOH = - log10[OH-] and pH = 14 – pOH.

When added to a buffer:

For eg, consider the buffer to be NaOAc/HOAc

HOAc is a weak acid and wont completely dissociate. So, the buffer solution has molecules of HOAc which will react with NaOH :

NaOH + HOAc = NaOAc + H2O

Thus, as NaOH is added to buffer, it gets neutralised, and no OH- is produced. So, the pH of buffer solution doesn’t change.

2.

To find Ka of a buffer solution, from the pH and concentration of salt and weak acid, use the Henderson equation :

[Salt] pH = pka + lograridi

Ka is acid dissociation constant.

Here, Ka corresponds to the acid dissociation of : HOAc → OAc- + H+

Here, salt = OAc- (NaOAc)

and acid = HOAc

[…] denotes concentration in molarity.

To find the concentrations :

First, the concentration from solution 3:

24.0 mL of 1M NaOAc is mixed with 10.0mL of 1 M HOAc.

So, the total final volume of the beaker =24.0 + 10.0 = 34.0 mL

The formula which is needed to calculate concentration after mixing/dilution:

V1S1 = V2S2

Where, V = Volume, S = Strength, and 1 and 2 denote initial and final solutions respectively.

(While using this formula, the units of Volume and Strength on both sides of the equation must be same.)

· For HOAc: the final solution has volume of 34.0 mL. So,

V2 = 34.0 mL, S2 = need to find.

The initial solution taken has a strength (S1) of 1M.

Volume of initial solution (V1) = 10.0 mL

Applying the formula :

10.0mL x 1M = 34.0 mL x S2

Solving, we get:

S2 = 0.294 M

· For NaOAc :the final solution has volume of 34.0 mL. So,

V2 = 34.0 mL, S2 = need to find.

The initial solution taken has a strength (S1) of 1M.

Volume of initial solution (V1) = 24.0 mL

Applying the formula :

24.0mL x 1M = 34.0 mL x S2

Solving, we get:

S2 = 0.706 M

These concentrations are further diluted in solution 4 by adding 40.0mL of deionised water to 10.0 mL of solution 3. So, calculating the final concentration using the same formula as above:

· For HOAc: the final solution has volume of 50.0 mL. So,

V2 = 50.0 mL, S2 = need to find.

The initial solution taken has a strength (S1) of 0.294M.

Volume of initial solution (V1) = 10.0 mL

Applying the formula :

10.0mL x 0.294M = 50.0 mL x S2

Solving, we get:

S2 = 0.0588 M

· For NaOAc :the final solution has volume of 50.0 mL. So,

V2 = 50.0 mL, S2 = need to find.

The initial solution taken has a strength (S1) of 0.706M.

Volume of initial solution (V1) = 10.0 mL

Applying the formula :

10.0mL x 0.706M = 50.0 mL x S2

Solving, we get:

S2 = 0.1412 M

Thus, finally, in solution 4, [NaOAc] = 0.1412 M and [HOAc] = 0.0588 M

And, pH of solution 4 = 4.98

Putting the values in Henderson equation:

[NaoAc] 4.98 = pKa + log [HOÀc]

0.1412 4.98 = pKa + logg

Solving, we get pKa = 4.6

Now, pKa = -log10(Ka)

So, 4.6 = -log10(Ka)

Or, 10-4.6 = Ka

Or, Ka = 2 x 10-5