Strong base is dissolved in 675 ml of 0.200 m weak acid (ka=3.25x10^-5) to make a buffer with a ph of 3.95. Assume that the volume remains constant when the base is added.
HA + OH ---> H2O + A^-
calculate the pka value of the acid and determine the number of moles of acid initially present
-so i got pka=4.488 and 0.135 mol HA
When the reaction is complete , what is the concentration ratio of conjugate base to acid.
i got [A^-]/[HA]=0.2897
Then it says , How many moles of strong base were initially added?... All i know the strong base has to be less than the number of moles of acid but i have no clue how to get the moles of OH^-.
Can you guys help me? Thank you so much!
See your post above. I didn't get the ratio to be 0.2897
is not 0.117
To calculate the number of moles of the strong base initially added (OH^-), you can use the pH of the solution and the dissociation constant (Ka) of the weak acid. Here are the steps to determine the moles of OH^- initially added:
1. Calculate the concentration of H+ ions in the solution using the formula pH = -log[H+]. In this case, the pH is given as 3.95, so [H+] = 10^(-3.95).
2. Determine the concentration of the weak acid (HA) initially present. This concentration is given in the problem as 0.200 M.
3. Use the Ka value to calculate the concentration of A^- ions in the solution. Since the reaction is HA + OH^- -> H2O + A^-, the concentration of A^- is equal to the concentration of OH^- when the reaction is complete.
Use the Ka expression: Ka = [H+][A^-] / [HA]. Since you know Ka (3.25x10^(-5)), [H+] (calculated in step 1), and [HA] (0.200 M), you can solve for [A^-]. Rearranging the expression, [A^-] = (Ka * [HA]) / [H+].
4. Now that you have the concentration of A^- ions, divide it by the concentration of HA to get the concentration ratio of conjugate base to acid: [A^-] / [HA]. In this case, you found it to be 0.2897.
The concentration ratio of [A^-] / [HA] tells you that for every 0.2897 moles of A^- initially present, there is 1 mole of HA initially present.
To find the moles of OH^- initially added, you can multiply the moles of HA initially present by the concentration ratio of [A^-] / [HA]:
Moles of OH^- = Moles of HA * [A^-] / [HA]
Moles of OH^- = 0.135 mol * 0.2897 = 0.0391 mol
Therefore, the moles of OH^- initially added is 0.0391 mol.