Question

You have a solution that contains the following acids at the given concentrations, what will be the pH of the solution?
[HF] = 0.02 M, pKa = 3.2
[HClO4] = 100 mL of a solution with 5.7 g of HClO4, pKa = ---
[HCl] = 100 mL of an acid 36.5% w / w and a density 1.81 g / mL 0.01 M, pKa = ----

Answer 1

HClO4 and HCl are strong acid nd ionizes completely.

a) Assumig density of HCl solutionn as 1.0 g/mL

Mass of HClO4 sollution = 100.0 g

Mass of HClO4 = 5.7 g

Mass of water = 100.0 - 5.7 = 94.3 g

Now, Maolar mass of HClO4 = 100.46 g//mol.

# of moles of HClO4 = Mass given / Molar mass = 5.7 g / 100.46 g/mol = 0.057 mole.

As HClO4 is strong acid,

# of mo^{les of H+ = # of moles of HClO4 = 0.057 mol.}

^{b) Molarity of HCl = 0.01 M and volume = 100.0 mL = 0.1 L}

^{# of moles of HCl = Molarity * Volume in L = 0.01 * 0.1 = 0.001 mole.}

^{As HCl a strong acid,}

^{# of moles of H+ = # of moles of HCl = 0.001 mol.}

^{c) HF is a weak acid and ionizes incompletely,}

HF (aq) H⁺ (aq) + F^- (aq)

Ka = [H⁺][F^-]/[HF]

pKa = 3.2

Ka = 10^-pKa = 10^-3.2 = 6.3*10^-4.

i.e. [H⁺][F^-]/[HF] = 6.3*10^-4. .......... (1)

Initially, [HF] = 0.02 M

let at equilibrim "X" M of HF ionized. The ICE table is,

HF (aq) H⁺ (aq) + F^- (aq)

Initially 0.02 M 0 0

Change -X +X +X

AT eqm. (0.02-X) X X

Using these equilibrium cocentrations in eq.(1),

(X)(X) / (0.02-X) = 6.3*10^-4.

X² / (0.02-X) = 6.3*10^-4.

As HF is weak acid we hav X << 0.02 and hence 0.02-X 0.02

X² / 0.02 = 6.3*10^-4.

X² = 0.02 * 6.3*10^-4.

X² = 12.6*10^-6.

Taking square root of both sides,

X = 3.55*10^-3 M

i.e. [H+] = 3.55*10^-3 M.

Now, total volume of the solution is 200.0 mL = 0.2 L

# of mole of H+ from HF = 3.55*10^-3 M * 0.2 L = 7.1 * 10^-4 mole.

Then,

Total # of moles of H+ = H⁺ from HF + H⁺ from HClO4 + H⁺ from HCl

= 7.1*10^-4 mol + 0.057 ml + 0.001 mol

= 0.05871 mol.

Volume of solution = 0.2 L

[H+] = # of moles / Volume in L = 0.05871 / 0.2 = 0.294 M

pH = -log([H+]) = -log(0.294)

pH = 0.532

pH of the solution will be 0.532

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