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
bob said it was something around 0.117 but i don't get that ..
3.95-4.488=-0.538
10^-0.538=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!
,
I worked this problem for you yesterday, in detail, and explained that 0.135 moles was right and 0.2897 for the ratio was not right. Why not look up your earlier post and you can get all of the details. Also I noted that you have written m. m stands for molality. M stands for molarity. I assumed you meant molarity.
To determine the number of moles of strong base initially added, we can use the equation for the reaction between the weak acid (HA) and the strong base (OH^-):
HA + OH^- → H2O + A^-
From the balanced equation, we can see that one mole of HA reacts with one mole of OH^- to produce one mole of A^-.
We previously calculated that the concentration of the weak acid was 0.200 M and the volume was 675 mL.
Therefore, the initial moles of acid (HA) can be calculated as:
moles of HA = concentration of HA x volume of HA
= 0.200 M x 0.675 L
= 0.135 moles
Since one mole of HA reacts with one mole of OH^-, the number of moles of OH^- initially added is also 0.135 moles.
To find the moles of the strong base (OH^-) initially added, you need to use the balanced chemical equation for the reaction:
HA + OH^- --> H2O + A^-
You already calculated that there are 0.135 mol of HA initially present. Since the reaction is 1:1, this means that there are also 0.135 mol of OH^- initially present.
Therefore, the moles of the strong base initially added is 0.135 mol.