ASAP need help am Stuck

Question

ASAP need help am Stuck

1. 8.0 mL of 0.10 M HC2H3O2 is titrated with 0.10 M NaOH. Perform the following calculations
2. pH of the acid before any NaOH is added 
3. Volume of 0.10 M NaOH needed to reach the equivalence point
4. pH at the equivalence point
5. Volume of 0.10 M NaOH needed to reach the half-way point
6. pH at the half-way point

I was stuck on this so could you show all steps for this please

Answer 1

Hello, Sdasd,

I will assume that "HC2H3O2" is acetic acid. It is a weak acid, so we need the equilibrium coefficient to determine how much of it dissociates into hydronium (H₃O⁺) and acetate ions (C₂H₃O^2-). I'll assume a K_a of 1.8x10^-5 Be sure to use the one given by the text, if not the one I used.

a. Make an ICE table showing what happens in the dissociation of acetic acid. The equation for the equilibrium is:

K_a = [H₃O⁺][C₂H₃O^2-]/[HC₂H₃O₂]

The balanced equation for this dissociation shows that we get 1 hydronium and 1 acetate ion for every 1 acetic acid. Set the concentrations of the hydronium and acetate both to x, since they are made from the same molecule in a ratio of 1 to 1. That means the concentration of the acetic acid will decrease by x, the same amount.

Update Note: Sdasd pointed out an error after I made an ICE table for this problem. I will correct that in this updated answer. Sorry, and thanks to Sdasd.

The change concerns the initial solution. I will assume the titration is complete, resulting in equimolar amounts of both the acetic acid and NaOH. Since they are the same initial concentration, there will be 8ml of each, resulting in 16 ml total volume. As a consequence, both initial concentrations of 0.10M are reduced to 0.05M.

Note: The ICE table won't fit in this response. It will appear in a second post.

Put the information into the equation, using the 0.05M for the concentration of the acid.:

1.8x10^-5 = x*x/(0.05-x)

The result is a quadratic equation. We can, and I will, take a shortcut by observing the the small equilibrium constant means not much acid will dissociate. That would mean x is small. If x is a lot smaller than the initial concentration, we could assume the quantity (0.05-x) is essentially the same as 0.05 and calculate the simper expression:

1.8x10^-5 = x²/(0.05)

x² = 9x10^-7

x = 9.49x10^-4 M

0.000949 M is small compared to 0.5 M, so we should be OK.

Let's check this by using the quadratic formula for a better answer.

-x² - 0.000018x + 9x10^-7 = 0

I find x = 0.000948M with the quadratic formula. Our simplifying assumption was reasonable, especially since we are limited to 2 sig figs with the starting numbers. I conclude the the simplified approach was acceptable, but not 100% accurate, in this problem.

Quadratic formula, x = 9.58x10^-4 or 9.6x10^-4 with 2 sig figs

First assumption: x = 9.49x10^-4 or 9.5x10^-4 with 2 sig figs

Assuming 0.00096M hydrogen ion, pH is defined as the negative log of the hydrogen ion concentration.

-log(0.00096) = 3.0, the pH

b. Volume of 0.10 M NaOH needed to reach the equivalence point can be found using moles and the relationship M1V1 = M2V2, where M and V are Molarity and volume of the acid(1) and base(2). This simplified approach can be used since one mole of acid reacts with 1 mole of base, a molar ratio of 1 to 1.

M1 = 0.10M, V1 = 8 ml

M2 = 0.10M, V2 = X, the unknown

(0.10M)*(8ml)=(0.1M)*X

X = 8 ml NaOH

c. pH at the equivalence point - I'm assuming it will be pH 7, the solution has as many OH^- as H⁺ ions, which presumably was the whole point (end point) of the titration.

d. I don't know what "half-way" point is. 8 ml/2 ????? Sorry.

e. Same problem for me as with (d). If this is when we add half of the 8ml of NaOH, then we could perhaps say that x/2 is (0.00096/2) or 0.00048 M H⁺. The negative log of this would be pH = 3.32

I became confused and made a wrong assumption in my first post, as caught by the poster, Sdas. I hope this response is correct, but be careful. Several assumptions were made in the process.

Bob