1.What is the pH of a 0.55M aqueous solution of hypobromous acid, HBrO, at 25?C
ID: 934326 • Letter: 1
Question
1.What is the pH of a 0.55M aqueous solution of hypobromous acid, HBrO, at 25?C (Ka=2.0x10-9) Answer) 4.48
2.What is the pH of a solution that contains 0.25M benzoic acid, HC7H5O2, and 0.75M sodium benzoate, NaC7H5O2? Ka(benzoic acid) = 6.3x10-5 Answer) 4.68
3.AlCl3 dissolves in water to a limited extent. If the measured molar solubility of AlCl3 in water is X, which of the expressions below most accurately represents the solubility product Ksp. Answer) 27X4
i have the answers i just need help solving these problems
Explanation / Answer
1)
HBrO <-> H+ and BrO-
K = [H+][BrO-]/[HBrO]
asume: [H+]=[BrO-] = x and [HBrO] = 0.55 (small error but neglectable
2*10^-9 = x^2 / 0.55
x = 3.3*10^-5
[H+] = 3.3*10^-5
pH= -log(3.3*10^-5) = 4.48
pH = 4.48
2)
HB: Benzoic Acid
NaB: Sodium Benzoate
HB <-> H+ and B-
NaB -> Na + and B-
There is a common ion, "B-", this is a buffer!
Buffer are best described with Henderson Hasselbach Equation
pH = pKa + log([salt]/[acid])
pKa = -log(Ka) = -log(6.3x10-5) = 4.2
Substitute
pH = 4.2 + log(0.75/0.25) = 4.2 + 0.477 = 4.677
pH = 4.68
3)
Limite extent means equilibrium in solubility!
AlCl3 <-> Al+3 and 3
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