Question

Describe how you would calculate the pH of the following 0.10 M aqueous solutions: (a) sodium monohydrogen phosphate (b) glycine hydrochloride (c) trisodium citrate For example: treat as monoprotic acid and use Kai; treat as intermediate form and use Ka, Ka... Find the pH of a solution prepared by dissolving 1.00 g of potassium hydrogen phthalate (204.221 g/mol) and 1.20 g of disodium phthalate (210.094 g/mol) in 50.0 mL of water. (pKai = 2.950, pK2 = 5.408) The diprotic acid has pKar = 4.00 and PK 2 = 8.00 (a) At what pH is [H2A] = [HA]? (b) At what pH is [HA] = [AP]? (c) Which is the principal (dominating) species at pH 2.00: H2A, HA', or A?-? (d) Which is the principal (dominating) species at pH 6.00? (e) Which is the principal (dominating) species at pH 10.00? Please use the appendix G from your textbook to find a conjugate acid/base pair that can be used to make a buffer solution of pH=6.80 and calculate the quotient between the two species.

thank you

Answer 1

1) a) Sodium monohydrogen phosphate has to be treated like an intermediate species between pKa₂ and PKa₃ of phosphoric acid.
b) Glycine hydrochloride has to be treated like a monoprotic acid with the Ka of the GlyH⁺ species (protonated glycine).
c) Trisodium citrate has to be treated as a monobase, with the Kb value associated to Ka₃ of citric acid. (remember that each pKa has an associated conjugate pKb value with the relation pKa + pKb = 14).

2) We have hydrogen phthalate and phthalate, which are related between them by the equlibrium:

$Hpht^{-}\leftrightharpoons pht^{2-}+H^{+}$

Which has the Ka₂ of phthalic acid as its equilibrium constant. This mixture is therefore a buffer, for which we can calculate the pH using the Henderson Hasselbach equation:

$pH=pKa_{2}+log(\frac{[pht^{2-}]}{[Hpht^{-}]})$

We can calculate their concentrations using, for potassium hydrogen phthalate:

$M=\frac{\frac{m}{m_{molar}}}{V(L)}=\frac{\frac{1.00g}{204.221g/mol}}{0.0500L}=0.0979M$

And for disodium phthalate:

$M=\frac{\frac{m}{m_{molar}}}{V(L)}=\frac{\frac{1.20g}{210.094g/mol}}{0.0500L}=0.114M$

So the pH is:

$pH=5.408+log(\frac{0.114M}{0.0979M})=5.475$

3) a) pKas can be defined as the pH values at which the concentration of the species involved in the equilibrium are equal (if you look at the Henderson-Hasselbach equation, you'll see that if the concentrations of the basic and acid species are the same, their ratio is 1, and log(1) = 0, so pH = pKa). This means that the answer here is pH = 4
b) At pH = pKa₂ = 8.00
c) At a pH vale below pKa₁, the dominating species is the most acidic; that is: H₂A.
d) Between pKa₁ and pKa₂, the dominating species is the intermediately basic one: HA^-.
e) At pH values above pKa₂, the predominating species is the most basic: A^2-.

4) I don't know which acids you have in your notebook, but the phosphoric acid system can be used, since its pKa₂ is 7.21, and we want the pKa value to be close to the desired pH value. The ratio can be calculated from the Henderson-Hasselbach equation:

$pH=pKa_{2}+log(\frac{[HPO_{4}^{2-}]}{[H_{2}PO_{4}^{-}]})$

(Bear in mind that these species are the ones involved in the second Ka os phosphoric acid!).

The ratio can be calculated as:

$\frac{[HPO_{4}^{2-}]}{[H_{2}PO_{4}^{-}]}=10^{pH-pKa_{2}}=10^{6.80-7.21}=0.389$