Question

A) Complete and balance the equation for each of the following reactions:

HI(aq)+Na2CO3(s) →

B) Ethylamine C2H5NH2), an organic amine and a weak base with Kb of 8.3×10−5, reacts with water to form its conjugate acid, C2H5NH3+.

Write the equation of dissociation for ethylamine.

C) A buffer solution is made by dissolving HC2H3O2 and NaC2H3O2 in water. Enter an equation that shows how this buffer neutralizes added acid .Enter an equation that shows how this buffer neutralizes added base. Calculate the pH of this buffer if the HC2H3O2 concentration is 0.24 M and the C2H3O2− concentration is 0.44 M . The Ka value for HC2H3O2 is 1.8×10−5.

Answer 1

A) A balanced reaction equation between HI (aq) and Na2CO3 (s) is,

2 HI (aq) + Na₂CO₃ (s) 2 NaI (aq) + H₂CO₃ (aq)

In presence of strong acid like HI, the H₂CO₃ (aq) will decompose into H₂O and CO₂ (g).

Hence Net Balanced equation is,

2 HI (aq) + Na₂CO₃ (s) 2 NaI (aq) + H₂O (l) + CO₂ (g).

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B)

C₂H₅NH₂ + H-OH   C₂H₅NH₃⁺(aq) + HO^- (aq) Kb =  8.3×10⁻⁵.

Equation of Kb is

Kb =  [C₂H₅NH₃⁺][HO^-]/ [C₂H₅NH₂] = 8.3×10⁻⁵.

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C) HC2H3O2 i.e. AcOH and NaC2H3O2 i.e. AcONa

AcOH-AcONa forms a buffer and resist change in pH i.e. [H⁺] and [HO^-]

i) Neutralization of Acid added say HA acid added,

HA acid is neutralized by AcO- ions in buffer furnished by dissociation of AcONa salt.

AcO^- (aq) + HA AcOH (aq) + A^- (aq)

ii) Neutralization of Base added say BOH base added,

BOH base is neutralized by AcOH acid in buffer which furnishes H+ ions.

AcO-H (aq) + BOH AcOB + H₂O

Hence the Neutralization of Acid & Base by buffer and resistance to pH change.

iii)

[AcOH] = 0.24 M, [AcO^-] = 0.44 M

Ka (AcOH) = 1.8×10⁻⁵. hence, pKa = -log Ka = -log(1.8×10⁻⁵) = 4.74

By Hendersons-Hasselbalch equation, pH of buffer is given as,

pH = pKa (AcOH) + log ([AcO^-] / [AcOH])

Placing known values,,

pH = 4.74 + log(0.44/0.24)

pH = 4.74 + 0.26

pH = 5.00

pH of the buffer solution is 5.00.

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