## Chemistry (4th Edition)

The $Base/Acid$ ratio in this buffer is equal to 10.55; therefore, it is more effective against added acids.
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: $K_a = \frac{[H_3O^+][HC{O_3}^-]}{[H_2CO_3]}$ $4.2 \times 10^{-7} = \frac{3.981 \times 10^{-8}*[HC{O_3}^-]}{[H_2CO_3]}$ $\frac{4.2 \times 10^{-7}}{3.981 \times 10^{-8}} = \frac{[HC{O_3}^-]}{[H_2CO_3]}$ $10.55 = \frac{[HC{O_3}^-]}{[H_2CO_3]}$ ------- 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.