pH = pKa + log (Ac^-)/(HAc)
Plug in pH, pKa, and solve for (Ac^-)/(HAc). That is equation 1.
Equation 2 is (HAc) + (Ac^-) = 0.05
Solve equations 1 and 2 simultaneously to find (Ac^-) and (HAc).
Then mol HAc = M x 0.250 = ?
and mols NaAc = M x 0.25 = ?
The above gives you the mols you will need which is what you asked for in the first part of the problem. I don't understand how the 0.1M solutions are to be used. Usually problems of this kind ask for mL of the solutions but this one doesn't do that so I ignored that part of the problem.
If you need to prepare 250.0 mL of a pH 5.00 buffer tha
t has a total buffer concentration of acetic acid +
sodium acetate of 0.050 M, how many moles of each will you need to prepare the solution? Given solutions
of acetic acid and sodium acetate with concentrations of 0.10 M and pKa= 4.76, describe how to pre
pare this
buffer.
4 answers
How do you solve those 2 equations simultaneously
That's math, not chemistry. :).
To make it easy typing let me call the acetic acid, a (for acid) and the sodium acetate, b (for base). So eqn 1 is pH = pKa + log b/a
eqn 2 is a+b = 0.05
Substitute into eqn 1 as follows:
5.00 = 4.76 + log b/a
5.00-4.76 = log b/a
0.24 = log b/a
b/a = 1.74 or
b = 1.74a
eqn 2 is a + b = 0.05
Substitute b from 1 into 2.
a + 1.74a = 0.05
2.74a = 0.05
a = 0.05/2.74 = ?
Then you substitute a back into eqn 2 of a + b = 0.05. Now you know the 0.05 and a, solve for b and go from there.
To make it easy typing let me call the acetic acid, a (for acid) and the sodium acetate, b (for base). So eqn 1 is pH = pKa + log b/a
eqn 2 is a+b = 0.05
Substitute into eqn 1 as follows:
5.00 = 4.76 + log b/a
5.00-4.76 = log b/a
0.24 = log b/a
b/a = 1.74 or
b = 1.74a
eqn 2 is a + b = 0.05
Substitute b from 1 into 2.
a + 1.74a = 0.05
2.74a = 0.05
a = 0.05/2.74 = ?
Then you substitute a back into eqn 2 of a + b = 0.05. Now you know the 0.05 and a, solve for b and go from there.
kindly give me hint for this
3 answers