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!
Note: m stands for molal. M stands for molar.
0.135 moles HA is correct.
4.88 is correct for pKa.
But (A^-)/(HA) = 0.117 and not your answer.
For the last part, how much OH was added initially.
.............HA + OH^- ==> A^- + H2O
initial....0.135..0.........0
added.............x
change.......-x..-x.........x......x
equil....0.135-x..0..........x......x
Since the 675 mL is the same for both base and acid, I can use the ratio simply as moles.
(A^-/HA) = 0.117
(x/0.135-x) = .117
Solve for x which will give you moles OH added as well as the moles base produced. I get something close to 0.014 (but you need to do it more accurately--and I like to check it to see if the ratio calculated really is 0.117).
how do you get 0.014 ? i seem to be getting another answer
How did you get the answer .117 ?
To find the moles of the strong base (OH⁻) initially added, we need to use the information given in the problem.
We know that the strong base reacts with the weak acid (HA) to form water (H₂O) and the conjugate base (A⁻). The balanced chemical equation for this reaction is:
HA + OH⁻ → H₂O + A⁻
From this equation, we can see that the stoichiometric ratio between HA and OH⁻ is 1:1. This means that for every mole of OH⁻ added, one mole of HA is consumed.
In the problem, we already found that the initial moles of HA present is 0.135 mol.
Since the neutralization reaction is complete and the volume remains constant, we can conclude that the moles of OH⁻ added are also 0.135 mol.
Therefore, the number of moles of strong base (OH⁻) initially added is 0.135 mol.