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lions [1.4K]
3 years ago
13

Will the ph of a solution of nh4cn be >7, <7, or =7?

Chemistry
1 answer:
never [62]3 years ago
7 0
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.


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What volume in mt, of 0.5a M1HCI solution is needed to neutralize 77 ml of 1.54 M NaOH solution?
Rainbow [258]

Answer:

237.2 mL.

Explanation:

  • We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.

(XMV) acid = (XMV) base.

where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.

M is the molarity of the acid or base.

V is the volume of the acid or base.

<em>(XMV) HCl = (XMV) NaOH.</em>

<em></em>

For HCl; X = 1, M = 0.5 M, V = ??? mL.

For NaOH, X = 1, M = 1.54 M, V = 77.0 mL.

<em>∴ V of HCl = (XMV) NaOH / (XV) HCl = (</em>1)(1.54 M)(77.0 mL) / (1)(0.5 M) = <em>237.2 mL.</em>

8 0
3 years ago
12.23 g of ammonia (NH3) are dissolved in enough water to make 560.0 mL of solution. How many moles of NH3 are added to the wate
Karolina [17]

Answer:

0.719 moles of NH₃

Explanation:

Molar mass of ammonia 17 g/mol

Mass of amonia = 12.23 g

Mass / Molar mass = Moles

12.23 g / 17 g/mol = 0.719 moles

4 0
3 years ago
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