Question

There are two sodium acetate solutions. Their apparent concentrations are 1.00 x 10^-1 M and 1.00 x 10^-10 M, respectively. Please find the pH and true concentration (M) of acetic acid, and sodium acetate in these two solutions, respectively. The pKa of acetic acid is 4.75.

Also, can you use the same indicator for the following titration? Why? Please reason your answers with at least two criteria. (a) HCl titrated by NaOH, and (b) NaOH titrated by HCl.

Answer 1

Solution :-

Solution 1

1.00*10^-1 M Sodium acetate

CH3COO- + H2O ----- > CH3COOH + OH-

0.1 M                                0                   0

-x                                       +x                 +x

0.1-x                                 x                    x

Kb = [CH3COOH][HO-]/[CH3COO-]

5.55*10^-10 = [x][x]/[0.10-x]

Since the kb is very small therefore neglect the x from the denominator

So we get

5.55*10^-10 = [x][x]/[0.10]

5.55*10^-10 * 0.10 = x^2

5.55*10^-11 = x^2

Taking square root of both sides we get

7.45*10^-6 = x =[OH-]

pOH = -log [OH-]

pOH= -log [ 7.45*10^-6]

pOH = 5.13

pH + pOH = 14

pH= 14 – pOH

pH = 14 – 5.13

pH= 8.87

Concentration of the [CH3COOH] = 7.45*10^-6 M

Concentration of the acetate = 0.10 – 7.45*10^-6 = 9.999*10^-2 M

Solution 2

1*10^-10 M acetate

CH3COO- + H2O ----- > CH3COOH + OH-

1*10^-10 M                                0                   0

-x                                       +x                 +x

1*10^-10-x                                 x                    x

Kb = [CH3COOH][HO-]/[CH3COO-]

5.55*10^-10 = [x][x]/[1*10^-10-x]

Since the kb is very small therefore neglect the x from the denominator

So we get

5.55*10^-10 = [x][x]/[1*10^-10]

5.55*10^-10 * 1*10^-10 = x^2

5.55*10^-20 = x^2

Taking square root of both sides we get

2.36*10^-10 = x =[OH-]

pOH = -log [OH-]

pOH= -log [ 2.36*10^-10]

pOH = 9.63

pH= 14 – pOH

pH= 14 – 9.63

pH= 4.37

concentration of the acetate = 1.0*10^-10 M

therefore the concentration of the acetic acid = 10^(-4.37) = 4.26*10^-5 M

part 2

For the titration of the HCl with NaOH and NaOH with HCl

We can use the same indicator because the color change is the importance of the indicator which can be seen either from color less solution to colored solution at the end point.

Or we can see the colored solution to the colorless solution at the end point depending on which titration is followed.