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:
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Answer:
Vol of 4 moles CO₂(g) at STP = 89.6 Liters
Explanation:
STP
P = 1 Atm
V =
T = 0°C = 273 K
n = 4 moles
R = 0.08206 L·Atm/mol·K
Using Ideal Gas Law PV = nRT => V = nRT/P
V = (4 moles)(0.08206 L·Atm/mol·K)(273 K)/(1 Atm) = 89.6 Liters
Answer:
13.75
Explanation:
Density is equal to mass divided by volume. 110/8 = 13.75
Answer:
The molar mass of the unknown acid is 386.8 g/mol
Explanation:
Step 1: Data given
Mass of the weak acid = 1.168 grams
volume of NaOH = 28.75 mL = 0.02875 L
Molarity of NaOH = 0.105 M
Since we only know 1 equivalence point, we suppose the acid is monoprotic
Step 2: Calculate moles NaOH
Moles NaOH = molarity NaOH * volume NaOH
Moles NaOH = 0.105 M * 0.02875 L
Moles NaOH = 0.00302 moles
We need 0.00302 moles of weak acid to neutralize the NaOH
Step 3: Calculate molar mass of weak acid
Molar mass = mass / moles
Molar mass = 1.168 grams / 0.00302 moles
Molar mass = 386.8 g/mol
The molar mass of the unknown acid is 386.8 g/mol
The answer is surface tension
