Sulfate buffer? your laboratory is out of materials to make phosphate buffer and you are considering using sulfate to make a buffer instead. the pka values for the two hydrogens in h2so4 are -10 and 2. (a) will this approach work for making a buffer effective near ph 7? (b) around what ph might a sulfate-based buffer be useful?

(a) No

(b) pH 2

Explanation:

A buffer is most useful when the amounts of the conjugate acid and conjugate base are in a 1:1 ratio. The Henderson-Hasselbalch equation can be used to relate the pH of a buffer to ratio of the conjugate acid-base pair:

pH = pKa + log([A⁻]/[HA])

To create a sulfate buffer at pH 7, the ratio of conjugate base to conjugate acid needs to be (using the second pKa value)

pH = pKa + log([A⁻]/[HA])

7 = 2 + log([A⁻]/[HA])

5 = log([A⁻]/[HA])

[A⁻]/[HA] = 10⁵

The ratio is nowhere near 1:1, so the buffer will not be effective.

(b) Since, a buffer is most effective at a 1:1 ratio of the conjugate acid-base pair, the pH should equal the pKa:

pH = pKa + log([A⁻]/[HA])

pH = pKa + log(1)

pH = pKa = 2

Thus, a sulfate-based buffer might be useful at pH 2

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number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.