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
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
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.
I apologize, but the answer provided is incorrect. Allow me to explain how to properly calculate the pH of the solution.
To find the pH of the solution, we first need to determine the concentration of the H+ ions in the solution. The titration reaction between NaOH and acetic acid can be represented as follows:
CH3COOH + NaOH -> CH3COONa + H2O
From the balanced equation, we can see that for every mole of acetic acid (CH3COOH) that reacts, one mole of H+ ions are produced. Therefore, the moles of H+ ions can be calculated using stoichiometry.
Given:
Volume of NaOH solution (VNaOH) = 25 mL = 0.025 L
Volume of acetic acid solution (VAcetic acid) = 21 mL = 0.021 L
Concentration of NaOH solution (CNaOH) = 0.45 mol/L
Concentration of acetic acid solution (CAcetic acid) = 0.35 mol/L
Using the formula:
moles of H+ ions = C x V
moles of H+ ions from NaOH = CNaOH x VNaOH
= 0.45 mol/L x 0.025 L
= 0.01125 mol
moles of H+ ions from acetic acid = CAcetic acid x VAcetic acid
= 0.35 mol/L x 0.021 L
= 0.00735 mol
Total moles of H+ ions = moles from NaOH + moles from acetic acid
= 0.01125 mol + 0.00735 mol
= 0.0186 mol
Now, we can calculate the molar concentration of the H+ ions in the solution:
Concentration of H+ ions = total moles of H+ ions / total volume of the solution
= 0.0186 mol / 0.046 L (since the total volume is 46 mL = 0.046 L)
= 0.4043 mol/L
Finally, to calculate the pH of the solution:
pH = -log[H+]
= -log(0.4043)
= 0.394
Therefore, the correct answer is pH = 0.394, rounded to three decimal places.