I have posted this twice and it doesn't want to post correctly. Here is another try. On the second HH equation it takes the I (initial) line and what follows onto the equation line of
base + H^+ ==> acid and it may do that again.
Use the HH equation with 3.5 for pH and solve for (base)/(acid) (which I will call b/a
b/a = ? and this is equation 1.
a + b = 0.05*4 is equation 2.
Solve these two equations simultaneously; a and b will be in mols.
Since your solution is 3.5 and you want the final solution to be 3.25, it must be more acidic; therefore, the equation you now need is
..........base + H^+ = acid
I.......above...0.....above
add..............x.............
C.........-x....-x.......+x
E.......above-x..0.......above+x
Substitute the E line into the HH equation and solve for x which is the mols H^+ that must be added.
Then M = mols/L. YOu know mols needed and M of the HCl, solve for L HCl needed.
I suggest you take the answer you get, prepare the solution on paper and work it to see that the pH really is 3.25.
Given stocks of 1M lactic Acid (pka 3.86) at pH 3.5, 2 M NaOH, 2M HCl and water; how would you prepare 4 Liters of 0.05M lactic acid at pH 3.25.
I know you use the Hendersen-Hasselbeck eqn, just not sure how to start. Please help.
2 answers
It appears to have posted correctly this time.