Calculate the volume of the acid solution and the volume of the conjugate base solution that would be needed to prepare a buffer with a total volume of 200. mL. Assume all stock solutions of the conjugate pair are 1.00 M. Show all of your work.

Explanation:

This can be contradictory, depending on whether the 0.1 M

is the total species concentration or the concentration of each of the two components. I'll consider this to be the former...

VA− = 9.125 mL

VHA = 15.875 mL

The Henderson-Hasselbalch equation is:

pH = pKa + log [A−][HA]

We have a pH 4.5

solution of acetic acid and acetate, so from there we can get the ratio of weak acid to conjugate base:

[A−][HA]=10

pH − pKa = 104.5 − 4.74 = 0.5754

Now, if the total concentration is

0.10 M , then:

[HA] + [A−] 0.5754

[HA] = 0.10 M

⇒[HA] = 0.10 M 1.0000 +0.5754

= 0.0635 M

−−−−−−−−

⇒[A−] = 0.0365 M

−−−−−−−−

and these concentrations are AFTER mixing. Since the total volume is 50 mL , or 0.050 L, the mols of each component (which are constant!) are:

nA − = 0.0365 molL × 0.050L =

0.001825 mols

−−−−−−−−−−−−

nHA = 0.0635 molL × 0.050L =

0.003175 mols

−−−−−−−−−−−−

So, if both of the starting concentrations were

0.20 M, we can find the volume they each start with:

VA − = 1 L0.20mols

A− × 0.001825mols A− = 0.009125 L = 9.125 mL

−−−−−−−−

VHA = 1 L 0.20 mols HA × 0.003175

mols HA = 0.015875 L = 15.875 mL

−−−−−−−−−

And this should make sense, because the total starting volume is

25.000 mL , the total ending volume is twice as large; the total species concentration is half the concentration that both species started with.

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.