To solve this problem, we need to analyze the enthalpy changes associated with the neutralization of the weak acids HA and HB with NaOH. We know the enthalpy of neutralization for each acid and the overall enthalpy change when both acids are present in the solution. Let's break this down step by step.
Understanding the Enthalpy Changes
The enthalpy of neutralization is the heat released when an acid reacts with a base to form water and a salt. In this case, we have:
- For acid HA: ΔHHA = -6900 cal/mol
- For acid HB: ΔHHB = -2900 cal/mol
When one mole of NaOH is added to a solution containing one mole of HA and one mole of HB, the overall enthalpy change observed is -3900 cal. This means that the reaction is exothermic, releasing heat.
Setting Up the Reaction Ratios
Let’s denote the amount of HA that reacts with NaOH as x moles and the amount of HB that reacts as y moles. According to the problem, the molar ratio of HA to HB is 1:T, which means:
Now, we can express the total enthalpy change for the reaction as follows:
Calculating Total Enthalpy Change
The total enthalpy change when one mole of NaOH is added can be expressed as:
ΔHtotal = (x * ΔHHA) + (y * ΔHHB)
Substituting the values we have:
-3900 cal = (1 * -6900 cal) + (T * -2900 cal)
Solving for T
Now, we can simplify this equation:
-3900 = -6900 - 2900T
Rearranging gives us:
-3900 + 6900 = -2900T
Which simplifies to:
3000 = -2900T
Now, dividing both sides by -2900:
T = -3000 / -2900
T = 3000 / 2900
Calculating this gives:
T ≈ 1.0345
Final Result
Thus, the value of T, which represents the molar ratio of the weak acid HB that reacts with NaOH compared to HA, is approximately 1.0345. This means that for every mole of HA that reacts, slightly more than one mole of HB reacts with NaOH, indicating that HB is less reactive compared to HA in this scenario.
