Answer:
pKa of the acid HA with given equilibrium concentrations is 6.8
Explanation:
The dissolution reaction is:
HA ⇔ H⁺ + A⁻
So at equilibrium, Ka is calculated as below
Ka = [H⁺] x [A⁻] / [HA] = 2.00 x 10⁻⁴ x 2.00 x 10⁻⁴ / 0.260
= 15.38 x 10⁻⁸
Hence, by definition,
pKa = -log(Ka) = - log(15.38 x 10⁻⁸) = 6.813
The theoretical proportion is given by the balanced chemical equation:
2 mol NBr / 3 mol Na OH
Then x mol NaOH / 40 mol NBr3 = 3mol NaOH/2 mol NBr3
Solve for x, x = 40 * 3/2 = 60 mol NaOH.
Given that there are 48 mol NaOH (less than 60) this is the limitant reactant and the other is the excess reactant.
Answer: NBr3..
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