Question

Saccharin is a weak organic base with a Kb of 4.80 × 10–3. A 0.297-g sample of saccharin dissolved in 25.0 mL of water has a pH of 12.190. What is the molar mass of saccharin? A sample of ammonia (Kb = 1.8 × 10–5) is titrated with 0.1 M HCl. At the equivalence point, what is the approximate pH of the solution? What is the equilibrium concentration of ammonium ion in a 0.33 M solution of ammonia (NH3, Kb = 1.8 × 10–5) at 25oC?

Answer 1

Saccharin is a weak organic base with a Kb of 4.80 × 10^–3.

A 0.297-g sample of saccharin dissolved in 25.0 mL of water has a pH of 12.190. What is the molar mass of saccharin?

1) From pH let us first find pOH and the [HO^-].

pH + pOH = 14.

pH = 12.190

pOH = 14 - 12.190

pOH = 1.81

[HO^-] = 10^-pOH. = 10^-1.81.

  [HO^-] = 10^-pOH. = 0.015.

Now a pertinent basicity equation of Saccharine(SN) and ICE table is,

Initial [SN] = X

SN + H-OH SNH⁺ (aq) + HO^- (aq)

Initially X 0 0

Change -0.015 + 0.015 + 0.015

Eqm conc. (X-0.015) 0.015 0.015

Kb = [SNH⁺][HO^-] / [SN]

Kb = 4.80 × 10^–3.

[SNH⁺][HO^-] / [SN] = 4.80 × 10^–3.

Using ICE table we have equilibrium concentrations in above equation,

0.015 x 0.015 / (X-0.015)= 4.80 × 10^–3.

X-0.015 = (0.015)² / (4.80 × 10^–3)

X - 0.015 = 0.047 M/L

X = 0.047 + 0.015

X = 0.062 M/L

Initial concentration of Sachharine is 0.062 M/L. i.e. 0.062 moles in 1L.

in 25 mL = 0.025 L we have,

# of moles of Sachharine = Molarity x volume = 0.062 x 0.025 = 0.00155 moles.

Mass of Sachharin taken = 0.297 g.

Now,

# of Moles = Mass taken / Molar mass of compound.

Molar mass of compound = Mass taken / # of moles.

Hence,

Molar mass of Saccharin = 0.297 / 0.00155 = 191.6 g/mole.

Molar mass of Saccharin is 191.6 g/mole.

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