Question

What is the [A-] when 50 ml of 0.1 M NaOH is added to the 100 ml solution mixture which is in 0.1 M HCl and 0.1 M acetic acid? (Ka =1.75*10-5) Lütfen birini seçin: a. 6.7*10-3M b. 1.32x10-3 M c. 3.6*10-6 d. 0.067 M e. 0.33 M

Answer 1

Answer:

[A⁻] = 0.067 M

Explanation:

In this case, let's see what happen in the mixture of the acid.

As we can see we have a strong acid and a weak acid, at the same concentration. so the moles would be the same:

moles HCl = moles CH₃COOH = 0.1 moles/L * 0.1 L = 0.01 moles

Now, the moles of the base would be:

moles NaOH = 0.1 * 0.050 = 0.005 moles

As the base is a strong base, it will react first with the strong acid, in this case, HCl as given:

            NaOH + HCl --------------> NaCl + H₂O

i.           0.005   0.010                      0         0

c.         -0.005  -0.005                 +0.005  +0.005

e.             0        0.005                  0.005    0.005

The base reacts completely with the strong acid, leaving the CH₃COOH completely in solution, so:

       CH₃COOH + NaOH -------------> CH₃COONa

i.          0.01               0                               0

e.            0                0                           0.01

So, the [A⁻] would be the [CH₃COONa]:

[A⁻] = 0.010 / (0.1 + .050)

[A⁻] = 0.067 M

Hope this helps