{"id":193756,"date":"2025-02-21T04:00:32","date_gmt":"2025-02-21T04:00:32","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=193756"},"modified":"2025-02-21T04:00:36","modified_gmt":"2025-02-21T04:00:36","slug":"consider-the-following-equilibrium-for-nitrous-acid-hno2-a-weak-acid","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/21\/consider-the-following-equilibrium-for-nitrous-acid-hno2-a-weak-acid\/","title":{"rendered":"Consider the following equilibrium for nitrous acid, HNO2, a weak acid"},"content":{"rendered":"\n

Consider the following equilibrium for nitrous acid, HNO2, a weak acid:<\/p>\n\n\n\n

HNO2 (aq) + H2O (l) <————–> H3O+ (aq) + NO2-<\/p>\n\n\n\n

In which direction will the equilibrium shift if<\/p>\n\n\n\n

NaOH is added?<\/p>\n\n\n\n

Sol18:<\/p>\n\n\n\n

When NaOH is added to the solution, it will react with HNO2 (the acid) to form NaNO2 (the salt) and water: HNO2 (aq) + NaOH (aq) ? NaNO2 (aq) + H2O (l)<\/p>\n\n\n\n

This reaction will remove HNO2 from the solution, which will shift the equilibrium to the left to try to restore the concentration of HNO2. As a result, more HNO2 will dissociate to form H3O+ and NO2- until a new equilibrium is reached.<\/p>\n\n\n\n

Therefore, the equilibrium will shift to the left if NaOH is added to the solution.<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

The explanation in your provided response contains an error regarding the direction of the equilibrium shift. When NaOH, a strong base, is added to a solution of HNO2, a weak acid, the reaction that occurs can be described as follows:<\/p>\n\n\n\n

[ \\text{HNO}_2 (aq) + \\text{NaOH} (aq) \\rightarrow \\text{NaNO}_2 (aq) + \\text{H}_2\\text{O} (l) ]<\/p>\n\n\n\n

In this reaction, NaOH completely dissociates into Na(^+) and OH(^-) ions. The OH(^-) ions from NaOH will react with the H(^+) ions produced by the dissociation of HNO2, forming water. This effectively reduces the concentration of H(^+) ions in the solution, which is represented by the decrease in H3O(^+).<\/p>\n\n\n\n

According to Le Ch\u00e2telier’s Principle, if the concentration of a product in a reversible reaction is decreased, the equilibrium will shift to the right to produce more of that product. In the case of the nitrous acid equilibrium:<\/p>\n\n\n\n

[ \\text{HNO}_2 (aq) + \\text{H}_2\\text{O} (l) \\rightleftharpoons \\text{H}_3\\text{O}^+ (aq) + \\text{NO}_2^- (aq) ]<\/p>\n\n\n\n

Adding NaOH decreases the concentration of H3O(^+) because the OH(^-) ions neutralize some of the H3O(^+). Consequently, to compensate for the loss of H3O(^+), the equilibrium will shift to the right, not to the left. This will lead to more HNO2 dissociating to form additional H3O(^+) and NO2(^-) until a new equilibrium is established.<\/p>\n\n\n\n

Therefore, the correct answer is that the equilibrium will shift to the right when NaOH is added to the solution, as this addition leads to the consumption of H3O(^+), and the system responds by producing more H3O(^+) and NO2(^-) to restore balance.<\/p>\n\n\n\n

To summarize:<\/p>\n\n\n\n