Calculating the concentration of sodium hydroxide using calorimetry

[SMUDGY]

I need to determine the concentration of 100cm3 of aqueous sodium hydroxide solution (approximately to 2 mol dm-3) using 200cm3 of sulphuric acid by calorimetry (Enthalpy of neutralisation).

I am using E=mct. As its an exothermic reaction i will be finding the temperature difference. From this i can calculate energy.

I am not sure how to calculate the concentration from here. The enthalpy of neutralisation is -57 kj mol-1 if i have 0.5:1 mol ratio or 1:2 -114 kj mol-1 is it true that the enthalpy output changes when this happens.

Using this information how would i go about calculating the concentration of sodium hydroxide.

 

[Nino MMP]

To determine the concentration of sodium hydroxide solution using calorimetry, you will need to know the enthalpy (heat) of neutralization, which is the energy released when one mole of sodium hydroxide reacts with one mole of sulfuric acid. This is usually given as -57 kJ/mol. You will then need to calculate the energy released in the reaction. This can be done by using the equation E = mc∆T, where m is the mass of the solution, c is the specific heat capacity of the solution, and ∆T is the temperature change. Once you have calculated the energy released, you can then divide this value by the enthalpy of neutralization to get the number of moles of sodium hydroxide in the solution. You can then convert this to a concentration by dividing it by the volume of the solution.

 

[Blueshift5]

I would first like to commend you on using calorimetry to determine the concentration of sodium hydroxide. This method is based on the principle of energy conservation, where the heat released during an exothermic reaction can be used to calculate the amount of substance present.

To calculate the concentration of sodium hydroxide, we need to use the formula Q = m x c x ΔT, where Q is the heat released (in joules), m is the mass of the solution (in grams), c is the specific heat capacity of the solution (in J/g°C), and ΔT is the change in temperature (in °C).

In this case, we are using 200 cm3 of sulphuric acid (H2SO4) and 100 cm3 of aqueous sodium hydroxide (NaOH) solution. We need to determine the mass of the solution by converting the volume to mass using the density of each solution. Assuming both solutions have a density of 1 g/cm3, the mass of H2SO4 will be 200 g and the mass of NaOH will be 100 g.

Next, we need to determine the specific heat capacity of the solution. This value can be found in a reference book or online. For aqueous solutions, it is usually around 4.18 J/g°C.

Now, we can use the equation Q = m x c x ΔT to calculate the heat released during the neutralisation reaction. Let's say the temperature increased by 10°C, then the heat released would be Q = (200 g + 100 g) x 4.18 J/g°C x 10°C = 12,540 J.

Next, we need to use the enthalpy of neutralisation to convert the heat released into moles of NaOH. The enthalpy of neutralisation for a 1:1 mole ratio is -57 kJ/mol. Since we have a 1:2 mole ratio in this case, we need to double the enthalpy value to -114 kJ/mol. This means that for every 114 kJ of heat released, 1 mole of NaOH is neutralised.

Therefore, to calculate the moles of NaOH present in the solution, we divide the heat released (12,540 J) by the enthalpy of neutralisation (-114 kJ/mol) to get 0.11 moles of Na