Question

Malonic acid (FW = 104.06) is a diprotic acid of the form CH2(COOH)2, with pKa1 = 2.847
and pKa2 = 5.696. Assume all solutions are ideal.
(a) If 15.000 g of malonic acid is added to 100 mL of pure water, what will the pH of the
solution be?
(b) You add 18 mL of 5 M NaOH to the solution in part (a). What will the new pH be?
(c) You then add 5.000 g of sodium malonate (Na2CH2(COO), FW = 148.025). What will
the new pH be?

Answer 1

Solution :-

Lets first calculate the moles of malonic acid

Moles = mass/ molar mass

           = 15.0 g / 104.06 g per mol

           = 0.1441 mol

Molarity of acid = 0.1441 mol / 0.1 L = 1.441 M

From the pka 1 and pka2 lets find the ka values

Ka = antilog [-pka]

Ka1 = antilog [-2.847]

      = 0.001422

Ka2 = antilog [-5.696]

       = 2.01*10^-6

Now lets calculate the H3O+ using the ka values assume CH2(COOH)2 = H2A

H2A + H2O -------- > H3O+ + HA^-

1.441                         0               0

-x                                   +x            +x

1.441-x                    x               x

Ka1 = [H3O+] [ HA-]/[H2A]

0.001422 = [x][x] /[1.441-x]

0.001422 * [1.441-x] = x^2

Solving for x we get

X=0.04456 M

Now lets use this x to find the H3O+ after second dissociation

HA- + H2O -------- > H3O+ + A^2-

0.04456                         0               0

-x                                   +x            +x

0.04456-x                    x               x

Ka2 = [H3O+][A^2-]/[HA^-]

2.01*10^-6 = [x][x]/[0.04456-x]

Since the ka2 is very small we can negelect the x from denominator tghen we get

2.01*10^-6 = [x][x]/[0.04456]

2.01*10^-6 * 0.04456 = x^2

8.96*10^-8 = x^2

Taking square root of both sides we get

2.98*10^-4 = x

So the total concentration of the H3O+ = 0.04456 + 2.98*10^-4 = 0.004486

1) Now lets calculate the initial pH

Initial pH= - log [H3O+]

                  = - log [0.04486]

                 = 1.35

   b) lets calculate moles of 18 ml of 5 M NaOH

moles of NaOH = 5 mol per L * 0.018 L = 0.09 mol NaOH

initial moles of acid = 0.1441 mol

so when we add NaOH then it reacts with acid to form conjugate base

so the moles of acid remain = 0.1441 mol – 0.09 mol = 0.0541 mol

moles of comnjugate base formed = 0.09 mol

new molarity at total volume is as follows

total volume = 100 ml + 18 ml = 118 ml = 0.118 L

[acid ] = 0.0541 mol / 0.118 L = 0.458 M

[conjuga base ] = 0.09 mol / 0.118 L = 0.763 M

Now lets calculate the pH

pH= pka1 + log [conj. Base]/[acid]

pH= 2.847 + log [0.763/0.458]

pH= 3.07

c) now lets calculate the pH after adding 5 g malonate

moles of malonate = 5.00 g / 148.05 g per mol = 0.03377 mol

new molarity of the malonate = 0.03377 mol / 0.100 L = 0.3377 M

pH = pka1 + log [base]/[acid]

pH= 2.847 + log [0.3377/1.441]

pH= 2.22