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Alexxx [7]
1 year ago
14

The NH₄⁺ ion forms acidic solutions, and the CH₃COO⁻ ion forms basic solutions. However, a solution of ammonium acetate is almos

t neutral. Do all of the ammonium salts of weak acids form neutral solutions? Explain your answer.
Chemistry
1 answer:
tigry1 [53]1 year ago
6 0

Not always ammonium salts of weak acids form neutral solutions.

When formic acid reacts with ammonia, ammonium formate is produced:

HCO2H + NH3 ---->  NH4HCO2

You already know that the weak conjugate bases of NH3 and HCO2H are NH4+ and HCO2, respectively.

How can the pH of the solution be calculated if the salt's anion causes the pH to rise and the salt's cation causes it to fall? The relative intensities of the basic anion and the acidic cation hold the key to the solution.

As was already established, formate is a weak base and will create hydroxide ions in water, whereas ammonium is a weak acid and will make hydronium ions in water.

NH4⁺ + H2O -----> NH3 + H3O⁺

HCO2⁻ + H2O ----->  HCO2H + OH⁻

Since the acid ionization of NH4+ is more favored than the base ionization of HCO2-, the solution will be acidic.

To learn more about ammonium salts:

brainly.com/question/10874844

#SPJ4

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If a gas occupies 79.5 mL at -1.4°C, what temperature, in Kelvin, would it
Anna35 [415]

Answer:

121 K

Explanation:

Step 1: Given data

  • Initial volume (V₁): 79.5 mL
  • Initial temperature (T₁): -1.4°C
  • Final volume (V₂): 35.3 mL

Step 2: Convert "-1.4°C" to Kelvin

We will use the following expression.

K = °C + 273.15 = -1.4°C + 273.15 = 271.8 K

Step 3: Calculate the final temperature of the gas (T₂)

Assuming ideal behavior and constant pressure, we can calculate the final temperature of the gas using Charles' law.

V₁/T₁ = V₂/T₂

T₂ = V₂ × T₁/V₁

T₂ = 35.3 mL × 271.8 K/79.5 mL = 121 K

5 0
2 years ago
If 507 g FeCl2 were used up in the reaction FeCl2 + 2NaOH Fe(OH)2(s) + 2NaCl, how many grams of NaCl would be made?
Ket [755]
Number of moles of FeCl2 used = mass/ molar mass 
Number of moles = 507/126.751 = 4.
If one mole of Fe reacts with two moles of sodium 
Then 4 moles of Fe produces 8 moles of sodium.
Number of moles of sodium = mass/molar mass
Molar mass of sodium chloride = 23 +35.5 = 58.5 g/mol
Hence mass = 8 * 58.5 = 468 g. Hence Option A.

6 0
3 years ago
Calculate the ph of a solution containing 0.0451 m potassium hydrogen tartrate and 0.028 m dipotassium tartrate. The ka values f
Marina86 [1]

Given buffer:

potassium hydrogen tartrate/dipotassium tartrate (KHC4H4O6/K2C4H4O6 )

[KHC4H4O6] = 0.0451 M

[K2C4H4O6] = 0.028 M

Ka1 = 9.2 *10^-4

Ka2 = 4.31*10^-5

Based on Henderson-Hasselbalch equation;

pH = pKa + log [conjugate base]/[acid]

where pka = -logKa

In this case we will use the ka corresponding to the deprotonation of the second proton i.e. ka2

pH = -log Ka2 + log [K2C4H4O6]/[KHC4H4O6]

     = -log (4.31*10^-5) + log [0.0451]/[0.028]

pH = 4.15



4 0
3 years ago
Answer this i will mark brainliest and don't steal points im gonna report you ​
Nuetrik [128]
The answer is for this is A
3 0
2 years ago
Read 2 more answers
QUESTION 1
Svetlanka [38]

Answer:

C

Explanation:

Because nothing is left at the end of the reaction.

3 0
3 years ago
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