40.0 mL of a 0.65 M solution of HF is titrated with 0.100 M NaOH. After 0.400 L of the NaOH solution has been added, the resultant solution is:

The solution is basic and after the equivalence point

Explanation:

HF reacts with NaOH as follows:

HF + NaOH → H₂O + NaF

1 mole of acid reacts per mole of base.

The solution of HF contains:

40.0mL = 0.0400L × (0.65mol / L) = 0.026 moles of HF,

The titration will reach equivalence point when you add 0.026 moles of NaOH. If more NaOH is added the solution will be basic because NaOH (A base) is in excess doing the solution basic.

Moles added of NaOH are:

0.400L × (0.100 mol / L) = 0.0400 moles of NaOH

As the addition of NaOH is > 0.026 moles,

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

the correct answer would be d. hope this !