Answer
The pH of blood containing these compounds is equal to $7.46$.
Work Step by Step
1. Drawing the ICE table, we get these concentrations at the equilibrium:
$CH_3COOH(aq) + H_2O(l) \lt -- \gt CH_3C{OO}^-(aq) + H_3O^+(aq)$
Remember: Reactants at equilibrium = Initial Concentration - x
And Products = Initial Concentration + x
$[CH_3COOH] = 2 \times 10^{-3} M - x$
$[CH_3C{OO}^-] = 0.025M + x$
$[H_3O^+] = 0 + x$
2. Calculate 'x' using the $K_a$ expression.
$ 4.3\times 10^{- 7} = \frac{[CH_3C{OO}^-][H_3O^+]}{[CH_3COOH]}$
$ 4.3\times 10^{- 7} = \frac{( 0.025 + x )* x}{ 2\times 10^{- 3} - x}$
Considering 'x' has a very small value.
$ 4.3\times 10^{- 7} = \frac{ 0.025 * x}{ 2\times 10^{- 3}}$
$ 4.3\times 10^{- 7} = 12.5x$
$\frac{ 4.3\times 10^{- 7}}{ 12.5} = x$
$x = 3.44\times 10^{- 8}$
Percent dissociation: $\frac{ 3.44\times 10^{- 8}}{ 2\times 10^{- 3}} \times 100\% = 1.72\times 10^{- 3}\%$
x = $[H_3O^+]$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 3.44 \times 10^{- 8})$
$pH = 7.46$
