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An acid is added to water, and a new equilibrium is established. What is the system after the acid is added?

A. [tex]\( pH \ \textless \ pOH \)[/tex] and [tex]\( K_w = 1 \times 10^{-14} \)[/tex]
B. [tex]\( pH \ \textless \ pOH \)[/tex] and [tex]\( K_w \ \textless \ 1 \times 10^{-14} \)[/tex]
C. [tex]\( pH \ \textgreater \ pOH \)[/tex] and [tex]\( K_w = 1 \times 10^{-14} \)[/tex]
D. [tex]\( pH \ \textgreater \ pOH \)[/tex] and [tex]\( K_w \ \textgreater \ 1 \times 10^{-14} \)[/tex]


Sagot :

When an acid is added to water, it increases the concentration of hydrogen ions ([tex]\( \text{H}^+ \)[/tex]) in the solution. This increase in hydrogen ions lowers the pH of the solution, making it more acidic (pH < 7).

In aqueous solutions at 25°C, the relationship between pH and pOH is given by:
[tex]\[ \text{pH} + \text{pOH} = 14 \][/tex]

Since the pH is less than 7 (due to the presence of the added acid), to satisfy the equation [tex]\(\text{pH} + \text{pOH} = 14\)[/tex], the pOH must be greater than 7. Therefore:
[tex]\[ \text{pH} < \text{pOH} \][/tex]

Another important constant in aqueous solutions is the ion-product constant for water ([tex]\( K_w \)[/tex]), which is given by:
[tex]\[ K_w = [\text{H}^+][\text{OH}^-] = 1 \times 10^{-14} \][/tex]

This constant ([tex]\( K_w \)[/tex]) remains the same at 25°C regardless of the changes in pH or pOH.

Given these observations, the correct outcome of the system after the acid is added is:
A. [tex]\( \text{pH} < \text{pOH} \)[/tex] and [tex]\( K _{ w } = 1 \times 10^{-14} \)[/tex]

Therefore, the answer is A.
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