Answer
The $Base/Acid$ ratio in this buffer is equal to 10.55; therefore, it is more effective against added acids.
Work Step by Step
1. Calculate $[H_3O^+]$:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 7.4}$
$[H_3O^+] = 3.981 \times 10^{- 8}M$
2. Write the $K_a$ equation, and find the ratio:
$K_a = \frac{[H_3O^+][HC{O_3}^-]}{[H_2CO_3]}$
$4.2 \times 10^{-7} = \frac{3.981 \times 10^{-8}*[HC{O_3}^-]}{[H_2CO_3]}$
$\frac{4.2 \times 10^{-7}}{3.981 \times 10^{-8}} = \frac{[HC{O_3}^-]}{[H_2CO_3]}$
$10.55 = \frac{[HC{O_3}^-]}{[H_2CO_3]}$
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Knowing that this ratio is greater than 1, the concentration of the base $(HC{O_3}^-)$ must be greater than $[H_2CO_3]$.
** This is mathematical: If the quocient is greater than 1, the dividend must be greater than the divisor.
Since the buffer has more base, added acids are neutralized more easily.
