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.
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.
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