A solution is prepared by dissolving 0.56 grams of benzoic acid (HC6H5CO2, MM = 122.12 g/mol, Ka = 6.4 X 10-5) in enough water to make 1.0 L of solution. Calculate the pH of this solution at equilibrium.
2. Calculate the pH of the system in Question 1 when 0.73 grams of soluble
KC6H5CO2 is added to the system. Assume the total solution volume is still 1.0 liters.
3. Calculate the pH change of the system in Question 2, if any, when 0.01 mol HBr is added to the system. Assume the total solution volume is still 1.0
liters.
4. Calculate the pH change of the system in Question 2, if any, when 0.01 mol KOH is added to the system. Assume the total solution volume is still 1.0 liters.
To solve these questions, we need to consider the dissociation of benzoic acid (HC6H5CO2) and the subsequent reactions with KC6H5CO2, HBr, and KOH.
1. The first step is to calculate the concentration of benzoic acid in the solution.
- The molar mass of benzoic acid (HC6H5CO2) is 122.12 g/mol.
- The given mass of benzoic acid is 0.56 grams.
- To convert grams to moles, divide the mass by the molar mass:
mole of benzoic acid = mass / molar mass = 0.56 g / 122.12 g/mol ≈ 0.00459 mol
- The final volume of the solution is 1.0 L, so the concentration of benzoic acid is:
concentration of benzoic acid = mole of benzoic acid / volume of solution = 0.00459 mol / 1.0 L = 0.00459 M
2. Next, we need to determine the concentration of KC6H5CO2 after it is added to the system.
- The given mass of KC6H5CO2 is 0.73 grams.
- The molar mass of KC6H5CO2 is 122.12 g/mol.
- Convert grams to moles:
mole of KC6H5CO2 = mass / molar mass = 0.73 g / 122.12 g/mol ≈ 0.00598 mol
- Since the total solution volume is still 1.0 L, the concentration of KC6H5CO2 is:
concentration of KC6H5CO2 = mole of KC6H5CO2 / volume of solution = 0.00598 mol / 1.0 L = 0.00598 M
3. To calculate the pH change when HBr is added, we need to consider the total moles of HBr added and the reaction with benzoic acid.
- The given moles of HBr is 0.01 mol.
- Since the total solution volume is still 1.0 L, the concentration of HBr is:
concentration of HBr = mole of HBr / volume of solution = 0.01 mol / 1.0 L = 0.01 M
- HBr will react with benzoic acid according to the equation: HBr + HC6H5CO2 ⇌ H3C6H5CO2 + Br-
- The reaction is a neutralization reaction, so it will consume an equal number of moles of H3C6H5CO2.
- The H3C6H5CO2 concentration change is equal to the HBr concentration: 0.01 M.
4. Finally, to calculate the pH change when KOH is added, we need to consider the total moles of KOH added and the reaction with benzoic acid.
- The given moles of KOH is 0.01 mol.
- Since the total solution volume is still 1.0 L, the concentration of KOH is:
concentration of KOH = mole of KOH / volume of solution = 0.01 mol / 1.0 L = 0.01 M
- KOH will react with benzoic acid according to the equation: KOH + HC6H5CO2 ⇌ KC6H5CO2 + H2O
- The reaction is a neutralization reaction, so it will consume an equal number of moles of HC6H5CO2.
- The HC6H5CO2 concentration change is equal to the KOH concentration: 0.01 M.
To calculate the pH of the solution and the pH changes when different substances are added, we need to walk through some steps.
1. Calculate the concentration of benzoic acid in the solution:
- Given mass of benzoic acid = 0.56 grams
- Molar mass of benzoic acid (HC6H5CO2) = 122.12 g/mol
- Convert mass to moles: moles = mass / molar mass
moles = 0.56 g / 122.12 g/mol = 0.00458 mol
- Since the solution is prepared in 1.0 L of water, the concentration of benzoic acid is: concentration = moles / volume (in liters)
concentration = 0.00458 mol / 1.0 L = 0.00458 M
2. Calculate the initial concentration of benzoate ions (C6H5CO2-) when KC6H5CO2 is added:
- Given mass of KC6H5CO2 = 0.73 grams
- Molar mass of KC6H5CO2 = 146.20 g/mol
- Convert mass to moles: moles = mass / molar mass
moles = 0.73 g / 146.20 g/mol = 0.00499 mol
- The concentration of KC6H5CO2 (which dissociates fully) is: concentration = moles / volume (in liters)
concentration = 0.00499 mol / 1.0 L = 0.00499 M
- Since KC6H5CO2 dissociates to form C6H5CO2- ions, the initial concentration of benzoate ions is also 0.00499 M.
3. Calculate the pH change when 0.01 mol HBr is added:
- Since HBr is a strong acid, it dissociates completely in water, contributing to the H+ ion concentration.
- The added 0.01 mol HBr increases the H+ ion concentration.
- To calculate the pH change, we need to know the volume of the solution and apply the formula: pH = -log[H+]
4. Calculate the pH change when 0.01 mol KOH is added:
- Since KOH is a strong base, it dissociates completely in water, contributing to the OH- ion concentration.
- The added 0.01 mol KOH increases the OH- ion concentration.
- To calculate the pH change, we need to know the volume of the solution and apply the formula: pH = 14 - log[OH-]
Note: The pH change depends on the initial values and the volume of the solution. Please provide the volume for accurate calculations.
You have posted four (four, count them) problems that are not easily done nor easily answered. I'll help with the first one and give some hints on the others but you need to do something along these lines yourself. Lets call benzoic acid HB..
(HB) - mols/L . mols = grams/molar mass = 0.56/122 appox 0.005 mols.Then mols/L = 0.005/1 L = about 0.005M.
..............HB ==> H^+ + B^-
I........0.005.........0..........0
C.........-x.............x...........x
E.....0.005-x.........x..........x
Write the Ka expression and substitute the E line into the Ka expression and solve for x = (H^+). Then pH = -log(H^+).
2. This is a buffered solution.Convert the K salt (the base) into M. Use the Henderson=Hasselalch equation. plug in base (the salt) and acid (benzoic acid and calculate the pH.
3 and 4. Set up an equation for reaction between the acid (HCl) or base (KOH) is added. For example 3 is
...................B^- + H^+ ==> HB
I..........
C...........
E..........You have concn of base to start (from the KB) and the amount of acid (HCl). Complete the ICE chart and use the HH equation to solve for pH.
Post your work if you get stuck.