Please let me know if my answer is correct.
Question:
A weak base with a concentration of 1.3 mol/L has a percent ionization of 0.72%. What is the Kb of this weak base?
Answer:
Calling the weak base SHO, SHO --> S^+ plus HO^-.
I know that the formula for Kb, therefore, will be Kb=(S^+)(HO^-)/SHO, After creating an ICE table, the initial concentration of SHO is equal to 1.3 mol/L.
H^+ is equal to zero and so is HO^- Regarding the change in concentration, S^+ is equal to +1.3 + 0.0072, HO^- is equal to +1.3 + 0.0072 and SOH^- is equal to 1.3 x 99.28.
Regarding Equilibrium concentration, S^+ is equal to 1.3 x 0.0072 =?, OH^- is equal to 1.3 x 0.0072 = ? and SOH is equal to 1.3 - (1.3 x 0.0072) or (1.3 x 0.9928 = ?.
After substituting the equilibrium numbers into Ka expression and calculating Ka,the step is written as follows: Ka=(S^+)(HO^-)/SHO^-
Subbing in the values gives Ka = (0.00936)(0.00936)/1.29064 gives the value of Ka a value of 6.8 x 10^-5
Please let me know if my answer to the following question is correct.
Question:
25 mL of standardized 0.45 mol/L NaOH is titrated with 21 mL of 0.35 mol/L acetic acid. Calculate the pH of the solution.
Answer:
Add 25 mL and 21 mL to get 46 mL.
Add 0.45 mol/L and 0.35 mol/L to get 0.8 mol/L.
Divide 0.0046 by 0.8 to get 5.8 x 10^-3.
Since pH = -log[H^+], pH = -log[5.8 x 10^-3] gives the answer to this question, which is a pH of 2.2
2 answers
The acetic acid/NaOH problem is not worked correctly. You can't add the volumes. Each neutralize each other. Try again.
The second problem, the one for Kb is worked correctly.
The second problem, the one for Kb is worked correctly.
1 answer