Question

Identify the species represented by each curve in the fractional composition diagram of a diprotic acid (H, A) with pK, values of 5.00 and 9.00. 1.00 0.90 0.80 0.70 Answer Bank Fractional composition A? HA HA 0 2 3 4 5 6 i 9 10 11 12 13 pH Calculate the fraction of the diprotic acid in each of its forms at pH 5.25. CH, A =

Answer 1

Part (a)

For a diprotic acid, dissociation is two step process which is as follows,

$H_2A\rightleftharpoons H^++HA^-\qquad (K_a_1)$

$H_A^-\rightleftharpoons H^++A^{2-}\qquad (K_a_2)$

Thus, for step 1, the equilibrium equation is given as,

$K_a_1=\frac{[H^+][HA^-]}{[H_2A]}\\ \Rightarrow -\log{K_a_1}=-\log{\frac{[H^+][HA^-]}{[H_2A]}}\qquad\textrm{[Taking negative log]}\\ \Rightarrow -\log{K_a_1}=-\log{[H^+]}-\log{\frac{[HA^-]}{[H_2A]}}\\ \Rightarrow pK_a_1=pH-\log{\frac{[HA^-]}{[H_2A]}}\qquad\textrm{(1)}$

Thus, when pH=pK_a1 = 5.00, we have, [HA^-] = [H₂A] i.e the curve of [HA^-] and [H₂A] are intersecting.

Similarly, for step 2, we will have,

$pK_a_2=pH-\log{\frac{[A^{2-}]}{[HA^-]}}\qquad\textrm{(2)}$

Thus, when pH=pK_a1 = 9.00, we have, [HA^-] = [A^2-] i.e the curve of [HA^-] and [A^2-] are intersecting.

Thus, we see that the curve of HA^- intersects twice, once the curve of H₂A at pH= 5.00 and then the curve of A^2- at pH=9.00. Hence this is the central curve (deep blue color).

The curve for H₂A intersects that of HA^- at pH=5.00. Thus, it is the red colored curve.

The third curve i.e. sky blue color one is of A^2-

Part (b)

Here, we will assume that [A^2-]=a, [HA^-] =b and [H₂A] =c.

pH=5.25

Thus, from Equation (1), we will obtain,

$pK_a_1=pH-\log{\frac{[HA^-]}{[H_2A]}}\\ \Rightarrow 5.00=5.25-\log{\frac{b}{c}}\qquad\textrm{[Putting the values]}\\ \Rightarrow \frac{b}{c}=10^{0.25}\\ \Rightarrow \frac{b}{c}=1.778\\ \Rightarrow c=0.562b\qquad (3)$

From equation (2), we have,

$pK_a_2=pH-\log{\frac{[A^{2-}]}{[HA^-]}}\\ \Rightarrow 9.00=5.25-\log{\frac{a}{b}}\qquad\textrm{[Putting the values]}\\ \Rightarrow \frac{a}{b}=10^{-3.75}\\ \Rightarrow a=1.778\times10^{-4}b\qquad (4)$

The fraction, $\alpha$ can be written as,

$\alpha_{H_2A}=\frac{c}{a+b+c}\\ \Rightarrow \alpha_{H_2A}=\frac{0.562b}{1.778\times10^{-4}b+b+0.562b}\qquad\textrm{[From (3 and (4))]}\\ \Rightarrow \alpha_{H_2A}\approx0.3599$

For HA^-

$\alpha_{HA^-}=\frac{b}{a+b+c}\\ \Rightarrow \alpha_{HA^-}=\frac{b}{1.778\times10^{-4}b+b+0.562b}\qquad\textrm{[From (3 and (4))]}\\ \Rightarrow \alpha_{HA^-}\approx0.6400$

For A^2-

$\alpha_{A^{2-}}=\frac{a}{a+b+c}\\ \Rightarrow \alpha_{A^{2-}}=\frac{1.778\times10^{-4}b}{1.778\times10^{-4}b+b+0.562b}\qquad\textrm{[From (3 and (4))]}\\ \Rightarrow \alpha_{A^{2-}}\approx0.000114$

The final answer is summarised in this figure

Identify the species represented by each curve in the fractional composition diagram of a diprotic acid (H2A) with pK of 5.00 and 9.00 values A2 Н.А НА DOC 1.00- 0.90- 0.80- 0.70 0.60 Аnswer Bank 0.50 Н.А НА 0.40 0.30 0.20- 0.10- 8 9 10 и 12 3 0 13 6 pH Calculate the fraction of the diprotic acid in each of its forms at pH 5.25 0.360 ан,А — 0.640 аНА 0.000114 ад Fractional composition