How many mols CH3COOH (HAc) do you have? That's M x L = 1.2 x 1.5 = 1.8
.........HAc + OH^- ==> Ac^- + H2O
I........1.8...0........0.......0
add............x..................
C.........-x..-x........+x......
E.......1.8-x...0........x
Now substitute the E line into the Henderson-Hasselbalch equation and solve for x = OH^ = mols OH to be added. I get 0.657 mols.
Then M NaOH = mols NaOH/L NaOH.
or L NaOH = mols NaOH/M NaOH
L NaOH = 0.657 mols/1.0 M = 0.657 L. :-)
What volume of 1.0 M NaOH must be added to 1.5 L of 1.20 M CH3COOH (ka=1.8x10^-5) to make the solution a buffer for pH 4.50?
I do not understand how to determine the mols of NaOH. I tried to use the equation where:
[H3O+] = ka (nHA/nA-)
but it is not giving me the correct answer of 0.657 L
5 answers
How do you get 0.657 mols? I solved for x and got 0.6529 mols
the preciseness in the original question is two sig digits...
however, I suspect your error is in the antilog of pH look there.
however, I suspect your error is in the antilog of pH look there.
Here is my work and I keep getting the hundredth value to be incorrect:
10^-4.5 = [1.8E-5 (1.8-x)]/x
10^-4.5 = [3.24E-5 - 1.8E-5]/x
10^-4.5x = 3.24E-5 - 1.8E-5x
4.96E-5x = 3.24E-5
x= 0.653
10^-4.5 = [1.8E-5 (1.8-x)]/x
10^-4.5 = [3.24E-5 - 1.8E-5]/x
10^-4.5x = 3.24E-5 - 1.8E-5x
4.96E-5x = 3.24E-5
x= 0.653
10^-4.5 = [1.8E-5 (1.8-x)]/x
10^-4.5 = [3.24E-5 - 1.8E-5x]/x
x(10^-4.5 +1.8E-5)=3.24E-5
x(10^.5+1.8)=3.24
x*4.96=3.24
x=.655
10^-4.5 = [3.24E-5 - 1.8E-5x]/x
x(10^-4.5 +1.8E-5)=3.24E-5
x(10^.5+1.8)=3.24
x*4.96=3.24
x=.655
1 answer