Consider the titration of a 24.0 mL sample of 0.105 M CH3COOH with 0.130 M NaOH. What is . . . (a) the initial pH? (b) the volume of added base required to reach the equivalence point? (c) the pH at 6.00 mL of added base? (d) the pH at one-half of the equivalence point? (e) the pH at the equivalence point?

See explanation below

Explanation:

First, we need to write the overall reaction which is:

CH₃COOH + NaOH <> CH₃COONa + H₂O

The acetic acid, is a weak acid, so it has a acid constant (Ka) which is 1.75x10⁻⁵. Now that we know this, we can solve the problem by parts:

a) Initial pH:

In this case, the base is not added yet, so the only thing we have is the acetic acid in solution. So we'll do an ICE chart for the dissociation of the acetic acid:

       CH₃COOH + H₂O <> CH₃COO⁻ + H₃O⁺     Ka  = 1.75x10⁻⁵

i)        0.105                                      0                0

e)       0.105 - x                                 x                x

Writting the Ka expression:    Ka = [CH₃COO⁻] [H₃O⁺] / [CH₃COOH]  replacing:

1.75x10⁻⁵ = x² / 0.105 - x

As Ka is a very small value, we can assume that x will be very small too, so we can assume that 0.105 - x = 0.105:

1.75x10⁻⁵ = x² / 0.105

1.75x10⁻⁵ * 0.105 = x²

x = [H₃O⁺] = 1.35x10⁻³ M

Then, the pH:

pH = -log[H₃O⁺]

pH = -log(1.35x10⁻³)

pH = 2.87

b) volume to reach the equivalence point

In this case, we use the following expression:

M₁V₁ = M₂V₂

And solve for the volume:

V₂ = 0.105 * 24 / 0.13

V₂ = 19.38 mL of NaOH

c) pH with 6 mL of added base

In this case we need to see the moles of each substance and then see how many moles of the acid remain:

moles Acid = 0.105 * 0.024 = 2.52x10⁻³  moles

moles base = 0.130 * 0.006 = 7.8x10⁻⁴ moles

remaining moles of acid: 2.52x10⁻³ - 7.8x10⁻⁴ = 1.74x10⁻³ moles

With these moles, we can calculate the concentration assuming the volume is the sum of the 24 mL and the 6 mL of added base (30 mL)

M = 1.74x10⁻³ / 0.030 = 0.058 M

Now we will do the same thing we did in part a) with an ICE chart to calculate the H₃O⁺ concentration and then, the pH.

      CH₃COOH + H₂O <> CH₃COO⁻ + H₃O⁺     Ka  = 1.75x10⁻⁵

i)        0.058                                      0                0

e)       0.058- x                                  x                x

1.75x10⁻⁵ = x² / 0.058

1.75x10⁻⁵ * 0.058 = x²

x = 1.01x10⁻³ M

pH = -log(1.01x10⁻³)

pH = 3

d) at one half equivalence point

In this case, if the volume to reach equivalence is 19.38 mL, the half would be 9.69 mL, and now we will do the same thing as part c) but with these data:

moles Acid = 0.105 * 0.024 = 2.52x10⁻³  moles

moles base = 0.130 * 0.00969 = 1.26x10⁻³ moles

remaining moles of acid: 2.52x10⁻³ - 1.26x10⁻³ = 1.26x10⁻³ moles

With these moles, we can calculate the concentration assuming the volume is the sum of the 24 mL and the 9.69 mL of added base (33.69 mL)

M = 1.26x10⁻³ / 0.03369 = 0.037 M

          CH₃COOH + H₂O <> CH₃COO⁻ + H₃O⁺     Ka  = 1.75x10⁻⁵

i)        0.037                                      0                0

e)       0.037- x                                  x                x

1.75x10⁻⁵ = x² / 0.037

1.75x10⁻⁵ * 0.037 = x²

x = 8.05x10⁻⁴ M

pH = -log(8.05x10⁻⁴)

pH = 3.09

e) pH at equivalence point.

In this case the moles of the acid and the base are the same, therefore the pH would be higher than 7. so in this case we need to use the following expression:

pH = 7 + 1/2 pKa + 1/2log[CH₃COOH]

In this case the concentration of the acid in the equivalence point would be:

[C] = 2.52x10⁻³ / 0.04338 = 0.058 M

Applying the above expression:

pH = 7 + 1/2(-log(1.75x10⁻⁵) + 1/2log(0.058)

pH = 8.76

answer: at room temperature (roughly 20 degrees c), you can dissolve 203.9 grams (or 200 grams) of sugar in 100 ml of water. as water temperature increases, the amount of sugar that will dissolve increases, up to about 500 grams of sugar in 100 ml water at 100 degrees c.

explanation:

it is the most commonly accepted one