PH of a solution will be <span>higher than 7
</span>
Ammonium cyanide is a salt formed by hydrogen cyanide and ammonia. Ammonia is a weak base and hydrogen cyanide is a weak acid.
NH₄CN + H₂O ⇒ NH₃ + HCN
NH₄⁺ + H₂O -----> H₃O⁺ + NH₃
CN⁻ + H₂O -----> HCN + OH⁻
Although both compounds are weak electrolytes, NH₃ is somewhat stronger base than HCN is a strong acid, so the solution reacts alkaline. We can prove this using Ka and Kb values:
Ka(HCN) = 4.9 x × 10⁻¹⁰
Kb(NH₃) = 1.8 × 10⁻⁵<span>
Kw= </span>1.0 × 10⁻¹⁴
Let's first calculate Ka for NH₄⁺:
Ka(NH₄⁺) x Kb(NH₃<span>) = pKw
</span>Ka(NH₄⁺) = Kw/Kb(NH₃) = 5.6 x 10⁻¹⁰
Then, Kb for CN⁻:
Kb(CN⁻) x Ka(HCN) = pKw
Kb(CN⁻) = Kw/Ka(HCN) = 2 x 10⁻⁵
From this, we can see that the acid constant NH4⁺ is much lower than the base constant of CN⁻, which will say that the solution of NH₄CN will react slightly alkaline because of the higher presence of hydroxyl ions in solution.
Answer:
18.22874999999973
I recommend you to round the nearest 1 d.p
Explanation:
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Answer:
S = 1.1 × 10⁻⁹ M
Explanation:
NaCl is a strong electrolyte that dissociates according to the following expression.
NaCl(aq) → Na⁺(aq) + Cl⁻(aq)
Given the concentration of NaCl is 0.15 M, the concentration of Cl⁻ will be 0.15 M.
We can find the molar solubility (S) of AgCl using an ICE chart.
AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)
I 0 0.15
C +S +S
E S 0.15+S
The solubility product (Ksp) is:
Ksp = 1.6 × 10⁻¹⁰ = [Ag⁺].[Cl⁻] = S (0.15 + S)
If we solve the quadratic equation, the positive result is S = 1.1 × 10⁻⁹ M
Answer:
Pure compounds include elements and compounds where a combination of two or more pure substances is a mixture. Only one type of atom comes in the form of a pure element. Multiple atoms consist of a molecule and different atoms consist of a compound. These are all pure substances and individually.
