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

[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)

Hope this helps

explanation:

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here in the venn diagram of sand and salt two circles represent sand and salt. but the area falling under the intersection of these circle represents the mixture of sand and salt.

so, in a venn diagram the mixture will be placed in a middle that is in between the intersection of two circles.

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