Sodium bicarbonate is often used to neutralize acids, such as the following reaction: NaHCO3 + HCl ---> NaCl + H20 + CO2.
What is the normality of a sodium bicarbonate solution containing 5.08 g NaHCO3 in 150.0 mL solution?
this chapter in chemistry is really confusing and i don't even know how to approach this problem. Please help
To determine the normality of a sodium bicarbonate solution, you need to follow a few steps:
1. Convert the given mass of NaHCO3 to moles.
2. Convert the volume of the solution to liters.
3. Calculate the molarity of the solution.
4. Determine the normality of the solution.
Let's go through each step in detail:
Step 1: Convert the mass of NaHCO3 to moles.
To do this, you need the molar mass of NaHCO3, which is 84.01 g/mol (sodium: 22.99 g/mol, hydrogen: 1.01 g/mol, carbon: 12.01 g/mol, oxygen: 16.00 g/mol).
moles of NaHCO3 = (mass of NaHCO3) / (molar mass of NaHCO3)
moles of NaHCO3 = 5.08 g / 84.01 g/mol
Step 2: Convert the volume of the solution to liters.
The given volume is 150.0 mL. To convert it to liters, divide by 1000:
volume (in liters) = 150.0 mL / 1000 = 0.150 L
Step 3: Calculate the molarity of the solution.
Molarity (M) is defined as moles of solute divided by liters of solution. In this case, the solute is NaHCO3.
Molarity (M) = moles of NaHCO3 / volume (in liters)
Molarity (M) = (moles of NaHCO3) / (0.150 L)
Step 4: Determine the normality of the solution.
In this reaction, 1 mole of NaHCO3 reacts with 1 mole of HCl, resulting in 1 equivalent of NaHCO3. Therefore, the normality of NaHCO3 is the same as its molarity.
Normality (N) = Molarity (M)
So, the normality of the sodium bicarbonate solution is the calculated molarity from Step 3.