The molal boiling point constant (Kb) for water is ).51oC/m and the molal freezing point constant (Kf) for water is 1.86oC/m. Given this information here is the question...
If I pour 25 mL of 0.035 M NaOH into 975 mL of water, what is the new concentration of NaOH?
You are analyzing a solid isolated from a reaction and find that its boiling point is over 1000 ∘C\rm ^{\circ}C. You dissolve the compound in water solution and recrystallize it to produce some solid crystals that you use to determine that the compound is a crystalline solid. Which of the following could be the identity of the unknown solid?
To find the new concentration of NaOH after adding it to water, we can use the concept of molality and the molal concentration constants (Kb and Kf) for water.
Step 1: Convert the volume of NaOH solution to mass:
First, we need to calculate the mass of NaOH in the given solution. To do this, we will use the molarity and volume of the NaOH solution. The molarity (M) represents the number of moles of solute (NaOH) per liter of solution.
Given:
Volume of NaOH solution (V) = 25 mL = 0.025 L
Molarity of NaOH solution (M) = 0.035 M
Mass of NaOH (m) = Molarity (M) x Volume (V)
= 0.035 M x 0.025 L
Step 2: Calculate the amount of solvent in kilograms:
The amount of solvent (water) is given in milliliters (mL) and needs to be converted to kilograms (kg) to calculate molality. The conversion factor is 1 mL = 1 gram = 0.001 kg.
Given:
Volume of water (Vw) = 975 mL = 975 g = 0.975 kg
Step 3: Calculate the molality of the solution:
Molality (m) is defined as the number of moles of solute (NaOH) per kilogram of solvent (water). It is calculated using the formula:
Molality (m) = Moles of solute (NaOH) / Kilograms of solvent (water)
First, let's calculate the number of moles of NaOH:
Moles of NaOH = Mass of NaOH / Molar mass of NaOH
The molar mass of NaOH can be calculated using the atomic masses:
Molar mass of Na = 22.99 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of NaOH = (22.99 g/mol) + (1.01 g/mol) + (16.00 g/mol)
= 40.00 g/mol
Moles of NaOH = (0.035 M x 0.025 L) / 40.00 g/mol
Next, let's calculate the molality:
Molality (m) = Moles of NaOH / Kilograms of water
= (0.035 M x 0.025 L) / (40.00 g/mol) / (0.975 kg)
Step 4: Calculate the change in boiling point (ΔTb) and freezing point (ΔTf):
The change in boiling point (ΔTb) and freezing point (ΔTf) can be calculated using the molality (m) and the molal constants (Kb and Kf) for water.
ΔTb = m x Kb
ΔTf = m x Kf
Given:
Kb (molal boiling point constant for water) = 0.51 oC/m
Kf (molal freezing point constant for water) = 1.86 oC/m
ΔTb = (molality) x (Kb)
ΔTf = (molality) x (Kf)
Step 5: Calculate the change in temperature:
The change in temperature can be positive or negative depending on whether it is a boiling point or freezing point. In this case, since we are dissolving NaOH in water, the freezing point will be lowered. Therefore, we will use the formula:
Change in temperature (ΔTf) = Original freezing point (0 oC) - New freezing point
New freezing point = Original freezing point - (ΔTf)
Step 6: Calculate the new concentration of NaOH:
Finally, to find the new concentration of NaOH, we need to add back the molality change (using the new freezing point) to the base concentration.
New concentration = (Change in moles of NaOH) / (Volume of solution + volume of water)
By following these steps, you should be able to calculate the new concentration of NaOH after pouring it into water.