The following solutions are combined in a calorimeter to determine the enthalpy of the neutralization reaction.
100.0 mL sodium hydroxide, NaOH, of unknown concentration an excess of 1.3732 M hydrochloric acid, HCl.
The experimental value for q is determined to be -2.676 kJ
Knowing that the theoretical value of q for any neutralization reaction is -55.90 kJ/mol H2O, determine the concentration of the sodium hydroxide solution.
NaOH + HCl ==> NaCl + H2O
mols NaOH = 0.100 L x M = ?. We need to find the M and we can do that if know the mols.
q for neutralization is 55.90 kJ/mol. We had q of -2.676 kJ so # mols in the reaction must have been -2.676/-55.90 = 0.04787
M NaOH = mols/L = 0.04787/0.1L = ? mols/L.
Well, let's get out our scientist hats and crunch some numbers! First things first, we need to convert the volume of the sodium hydroxide solution to moles.
To do that, we'll use the molarity (M) formula: M = moles / liters. Since we know the molarity of the hydrochloric acid is 1.3732 M and we have an excess of it, we can assume that the volume of the hydrochloric acid is equal to the volume of the sodium hydroxide solution.
So, the volume of the sodium hydroxide solution is 100.0 mL. We convert that to liters by dividing by 1000, giving us 0.1000 L.
Now, to find the moles of sodium hydroxide, we'll use the molarity formula:
Moles = Molarity * Volume
Moles = (Unknown Concentration) * 0.1000 L
Now, from there we can find the mols of water produced in the reaction because, well, you can't have a neutralization reaction without water!
The theoretical value of q for any neutralization reaction is -55.90 kJ/mol H2O. We have the experimental value of q, which is -2.676 kJ. And since we know that 1 mol of water is produced in this reaction, we can set up a proportion:
-2.676 kJ / -55.90 kJ = 1 mol H2O / x mol H2O
Solving for x, we find that x is equal to 0.04791 mol H2O.
Now, since we have the moles of sodium hydroxide and the moles of water, we can determine the concentration of the sodium hydroxide solution.
Molarity = Moles / Volume
Molarity = 0.04791 mol H2O / 0.1000 L
Drumroll, please! The concentration of the sodium hydroxide solution is approximately 0.4791 M.
And there you have it! The concentration of the sodium hydroxide solution is approximately 0.4791 M. I hope those calculations didn't neutralize your enthusiasm!
To determine the concentration of the sodium hydroxide (NaOH) solution, we can use the equation:
q = mCΔT
Where:
q = heat transfer (in joules or kilojoules)
m = mass of the solution (in grams)
C = specific heat capacity of the solution (assumed to be 4.18 J/g·°C for water-based solutions)
ΔT = change in temperature (in °C)
In this case, the heat transfer (q) is given as -2.676 kJ (-2.676 × 10^3 J), and we need to find the concentration of NaOH. We know that the theoretical value of q for any neutralization reaction is -55.90 kJ/mol H2O.
First, we need to convert the volume of NaOH solution (100.0 mL) to grams, assuming the density of the solution is 1 g/mL.
Step 1: Convert the volume to grams:
100.0 mL × 1 g/mL = 100.0 g
Next, we need to calculate the moles of water (H2O) produced in the reaction, using the molar enthalpy of -55.90 kJ/mol H2O.
Step 2: Calculate moles of H2O:
Moles of H2O = q / ΔH
Moles of H2O = -2.676 × 10^3 J / (-55.90 kJ/mol) = -2.676 × 10^3 J / (-55.90 × 10^3 J/mol) = 0.0478 mol
Since the balanced equation for neutralization of NaOH with HCl is:
NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l)
We can see that the mole ratio between NaOH and H2O is 1:1. Therefore, the moles of NaOH is also 0.0478 mol.
Finally, we can determine the concentration of NaOH in the solution:
Step 3: Calculate the concentration of NaOH:
Concentration = moles / volume
Concentration = 0.0478 mol / 0.100 L = 0.478 mol/L
So, the concentration of the sodium hydroxide solution is 0.478 M.
To determine the concentration of the sodium hydroxide (NaOH) solution, we can use the experimental value of q (heat) and the theoretical value of q for the neutralization reaction.
First, let's calculate the number of moles of water (H2O) involved in the neutralization reaction. From the balanced equation for the neutralization reaction between NaOH and HCl:
NaOH + HCl → NaCl + H2O
We know that the theoretical value of q for the neutralization reaction is -55.90 kJ/mol H2O. Since we have the experimental value for q as -2.676 kJ, we can calculate the number of moles of water:
moles H2O = (experimental value of q) / (theoretical value of q)
moles H2O = -2.676 kJ / -55.90 kJ/mol = 0.0478 mol H2O
Next, we need to determine the number of moles of NaOH that reacted. From the balanced equation, we know that 1 mole of NaOH reacts with 1 mole of H2O. Therefore, the number of moles of NaOH is also 0.0478 mol.
Now we can calculate the concentration of the sodium hydroxide solution. The concentration is given in terms of moles per liter (M). We have 100.0 mL of NaOH solution, which is equivalent to 0.1000 L.
concentration of NaOH = (moles NaOH) / (volume of NaOH solution in liters)
concentration of NaOH = 0.0478 mol / 0.1000 L = 0.478 M
Therefore, the concentration of the sodium hydroxide solution is 0.478 M.