Latihan Soal Hidrolisis garam II Kimia Kelas XI

Takeaways

• 😀 The first part of the script explains the process of determining the pH of a mixture of H2CO3 and CaOH2, using stoichiometric calculations and concentration formulas.
• 😀 The reaction between H2CO3 and CaOH2 results in the formation of CaCO3 and H2O, with no remaining H2CO3 or CaOH2 in the solution after the reaction is complete.
• 😀 The concentration of CaCO3 in the final solution is calculated by dividing the number of moles of CaCO3 by the total volume, yielding a concentration of 0.1 M.
• 😀 The script explains how to determine the OH- concentration using the formula: sqrt(Kw / Kb * [anion]), where the anion in the case of CaCO3 is CO32-.
• 😀 The OH- concentration for the reaction is calculated as 5 x 10^-5 M, which leads to the calculation of pOH and then pH for the final solution.
• 😀 In the second example, H2SO4 reacts with NH4OH to form NH4SO4, with the pH of the final solution being 6. The script calculates the necessary concentrations of H2SO4 and NH4OH to achieve this pH.
• 😀 For the H2SO4 and NH4OH reaction, the H+ concentration is determined from the given pH, and then the NH4+ concentration is calculated using the formula for base dissociation.
• 😀 The concentration of NH4+ is found to be 2 x 10^-3 M, and the corresponding molar concentrations of H2SO4 and NH4OH are determined to be 4 x 10^-3 M and 8 x 10^-3 M, respectively.
• 😀 In the third example, the script calculates the pH of a solution of magnesium cyanide (Mg(CN)2) by first determining its molarity and then finding the OH- concentration using a similar formula as in the previous examples.
• 😀 The script calculates the OH- concentration from the dissociation of cyanide ions (CN-) and finds that the pH of the solution is 11 + log(2).
• 😀 In the fourth example, the script explains how to find the mass of calcium oxalate (CaC2O4) in a 1-liter solution, using stoichiometric calculations and the formula for pH, involving the Ka of oxalic acid and the resulting OH- concentration.
• 😀 The script concludes with the calculation of the molar mass of calcium oxalate, leading to the determination of its mass in the solution, which is 768 grams.

Q & A

What is the first problem discussed in the transcript about hydrolysis?

The first problem discusses the hydrolysis of a mixture of H2CO3 and Ca(OH)2, where the goal is to determine the pH of the resulting solution. The concentrations and volumes of the reactants are given, along with the dissociation constant for H2CO3 (Ka = 4 × 10⁻⁷).

How is the concentration of CaCO3 determined in the first problem?

The concentration of CaCO3 is determined by dividing the number of moles (20 mmol) by the final volume (200 mL), resulting in a concentration of 0.1 M.

What equation is used to calculate the concentration of OH⁻ ions in the first problem?

The concentration of OH⁻ ions is calculated using the equation: [OH⁻] = √(Kw / Kb × [anion]), where Kw is 10⁻¹⁴, Kb is 4 × 10⁻⁷, and the anion concentration is 0.1 M.

What is the significance of the pH being determined from the OH⁻ concentration?

Since the solution contains a strong base (Ca(OH)2) and a weak acid (H2CO3), the concentration of OH⁻ ions is used to find the pH. The pH is then derived by calculating pOH and subtracting it from 14 (the total of pH + pOH).

What happens in the second problem involving H2SO4 and NH4OH?

In the second problem, H2SO4 reacts with NH4OH to form ammonium sulfate (NH4)2SO4, with a given pH of 6. The goal is to determine the concentrations of H2SO4 and NH4OH that were used in the reaction.

How is the concentration of NH4⁺ ions calculated in the second problem?

The concentration of NH4⁺ is calculated by solving the equation for H⁺ concentration: [H⁺] = √(Kw / Kb × [NH4⁺]), with known values for Kw (10⁻¹⁴) and Kb (2 × 10⁻⁵), resulting in a concentration of NH4⁺ equal to 2 × 10⁻³ M.

How do we calculate the concentration of H2SO4 used in the second problem?

The concentration of H2SO4 is calculated by first determining the amount of NH4⁺ formed and using stoichiometry to find the corresponding molar amount of H2SO4, which is found to be 4 × 10⁻³ M.

What is the third problem about, and what is the main calculation performed?

The third problem involves the dissolution of magnesium cyanide (Mg(CN)2) in water. The goal is to find the pH of the resulting solution, which involves calculating the concentration of OH⁻ ions based on the dissociation of cyanide ions and the equilibrium constant (Ka) for HCN.

How is the concentration of OH⁻ ions calculated in the third problem?

The concentration of OH⁻ is calculated using the equation: [OH⁻] = √(Kw / Kb × [cyanide]), where Kb for cyanide is derived from the given Ka of HCN (6 × 10⁻¹⁰) and the concentration of cyanide is calculated based on the molar mass of magnesium cyanide.

What is the purpose of using Ka and Kb in these problems?

Ka (acid dissociation constant) and Kb (base dissociation constant) are used to determine the extent of dissociation of weak acids and bases in solution. By knowing these constants, we can calculate the concentrations of ions such as H⁺ and OH⁻, which are crucial for determining the pH of the solution.