Question

For the diprotic weak acid H₂A, ?a1=2.9×10−6 and ?a2=8.4×10−9

What is the pH of a 0.07000.0700 M solution of H2A?

What are the equilibrium concentrations of H2A and A2− in this solution?

Answer 1

Answer:-

1. $p^{H} = 3.35$.

2. equilibrium [H₂A] = 0.0695409 M

3. equilibrium [A^2-] = $8.4 \times 10^{-9} M$

Explanation :-

Given:-

$Molarity Of H_{2}A = 0.0700 M$

$K_{a1} = 2.9 \times 10^{-6}$

$K_{a2} = 8.4 \times 10^{-9}$

Now, First dissociation is

$H_{2}A \rightleftharpoons H^{+} + HA^{-}$ ...(1)

The ICE Table is

	$H_{2}A$	$\rightleftharpoons$	$H^{+}$	+	$HA^{-}$
initial	0.07		0		0
change	-x		+x		+x
equilibrium	0.07 -x		x		x

Now, from equilibrium (1) we have,

$K_{a1} = \frac{[H^{+}][HA^{-}]}{[H_{2}A]}$

$2.9 \times 10^{-6} = \frac{x.x}{0.07 -x}$

but, x << 0.07 hence, $0.07 -x \approx 0.07$

therefore,

$2.9 \times 10^{-6} = \frac{x^{2}}{0.07}$

ie. $x^{2} = 2.9 \times 10^{-6} \times 0.07$

$x^{2} = 0.203 \times 10^{-6}$

$x = 0.451 \times 10^{-3} = 4.51 \times 10^{-4}$

ie. [H⁺] = [HA^-] = $4.51 \times 10^{-4}$

Now second dissociation is

$HA^{-} \rightleftharpoons H^{&plus;} &plus; A^{2-}$...(2)

The ICE table is

	$HA^{-}$	$\rightleftharpoons$	$H^{+}$	+	$A^{2-}$
initial	$4.51 \times 10^{-4}$		$4.51 \times 10^{-4}$		0
change	-x		$4.51 \times 10^{-4} + x$		+x
equilibrium	$4.51 \times 10^{-4} - x$		$4.51 \times 10^{-4} + x$		x

Now from equilibrium (2) we have,

$K_{a2} = \frac{[H^{&plus;}][A^{2-}]}{[HA^{-}]}$

ie. $8.4 \times 10^{-9} = \frac{(4.51 \times 10^{-4} &plus;x)(x)}{(4.51 \times 10^{-4} - x)}$

but x is very small

therefore,

$8.4 \times 10^{-9} = \frac{(4.51 \times 10^{-4})(x)}{(4.51 \times 10^{-4})}$  

ie. $8.4 \times 10^{-9} = x$

Therefore, [H⁺] = $(4.51 \times 10^{-4} &plus; 8.4 \times 10^{-9}) \approx 4.51 \times 10^{-4} M$

Now,

$p^{H} = - log [H^{&plus;}]$

$p^{H} = - log (4.51 \times 10^{-4})$

$p^{H} = - (-3.35)$

$p^{H} = 3.35$

Again,

$equilibrium [H_{2}A] = 0.07 - x$

$equilibrium [H_{2}A] = 0.07 - 4.51 \times 10^{-4} = 0.069549 M$  

and

equilibrium [A^2-] = $8.4 \times 10^{-9} M$