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:
Scandium is the smallest element of the four
Explanation:
Vanadium = 50.9415 u
Titanium = 47.867 u
Scandium = 44.955912 u
Chromium = 51.9961 u
Answer:
Single Displacement reaction
In a displacement reaction, a more reactive element replaces a less reactive element from a compound.
Change in colour takes place with no precipitate forms.
Metals react with the salt solution of another metal.
Examples:
2KI + Cl2 → 2KCl + I2
CuSO4 + Zn → ZnSO4 + Cu
Double displacement reaction
In a double displacement reaction, two atoms or a group of atoms switch places to form new compounds.
Precipitate is formed.
Salt solutions of two different metals react with each other.
Examples:
Na2SO4 + BaCl2 → BaSO4 + 2NaCl
2KBr + BaCl2 → 2KCl + BaBr2
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Explanation:
The balanced equation of the reaction is given as;
Mg(OH)2 (s) + 2 HBr (aq) → MgBr2 (aq) + 2 H2O (l)
1. How many grams of MgBr2 will be produced from 18.3 grams of HBr?
From the reaction;
2 mol of HBr produces 1 mol of MgBr2
Converting to masses using;
Mass = Number of moles * Molar mass
Molar mass of HBr = 80.91 g/mol
Molar mass of MgBr2 = 184.113 g/mol
This means;
(2 * 80.91 = 161.82g) of HBr produces (1 * 184.113 = 184.113g) MgBr2
18.3g would produce x
161.82 = 184.113
18.3 = x
x = (184.113 * 18.3 ) / 161.82 = 20.8 g
2. How many moles of H2O will be produced from 18.3 grams of HBr?
Converting the mass to mol;
Number of moles = Mass / Molar mass = 18.3 / 80.91 = 0.226 mol
From the reaction;
2 mol of HBr produces 2 mol of H2O
0.226 mol would produce x
2 =2
0.226 = x
x = 0.226 * 2 / 2 = 0.226 mol
3. How many grams of Mg(OH)2 are needed to completely react with 18.3 grams of HBr?
From the reaction;
2 mol of HBr reacts with 1 mol of Mg(OH)2
18.3g of HBr = 0.226 mol
2 = 1
0.226 = x
x = 0.226 * 1 /2
x = 0.113 mol
