Question

${\huge{\tt Questions}}$

1.The ionization constant of a base ${(C_2H_5)_3N}$ is 6.4 × 10^(-5) .calculate the degree of dissociation and 0.1 M solution when it is mixed with 0.01 M NaOH solution.

$\rule{300}{1.5}$

2. calculate the pH of a 0.033 m Ammonia solution is 0.33 M ${NH_4Cl}$ is introduced in the solution at the same temperature .
${(k_b \: For \: NH_3 = 1.77 \times {10}^{ - 5})}$

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Answer 1

Answer:

1. 6.4 × 10-³

2. pH of the solution = 9.25

Explanation:

1. Degree of dissociation = Ka / Concentration of strong electrolyte having common ion added

= (6.4 × 10-⁵) / 0.01

= 6.4 × 10‐³

Degree of dissociation of (C₂H₅)₃N is 6.4 × 10-³

2. To calculate the pH of the solution containing ammonia and NH4Cl, we will use the Handerson Hasselbach equation.

According to this equation,

pOH = pKb + log [Salt]/[Base]

where,   pKb is to be calculated from Kb.

Kb for NH3 = 1.77×10-5

pKb = -log Kb = -log (1.77×10-5) = 4.75

[Salt] = [NH4Cl] = 0.033 M

[Base] = [NH3] = 0.033 M

When we put all these values in the equation,

pOH = 4.75 + log 0.0330.033

= 4.75 + log (1) = 4.75+0 = 4.75

pOH = 4.75

pH = 14 - 4.75 = 9.25

pH = 9.25

pH of the solution = 9.25

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Answer 2

Answer:

This is the answer for the question sir