Homework Help: chemistry

Posted by Abigail on Tuesday, April 15, 2008 at 4:55pm.

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

chemistry - Abigail, Tuesday, April 15, 2008 at 4:58pm
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
chemistry - DrBob222, Tuesday, April 15, 2008 at 5:25pm
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.

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