2.00g of NaOH are dissolved in water to make 2.00L of solution. What is the concentration of hydronium ions, [H3O+], in this solution?
To determine the concentration of hydronium ions, [H3O+], in the solution, we first need to calculate the number of moles of NaOH in 2.00g of the compound.
The molar mass of NaOH is:
Na (sodium) + O (oxygen) + H (hydrogen) = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 39.99 g/mol
Using the molar mass, we can calculate the number of moles of NaOH:
moles = mass / molar mass
moles = 2.00g / 39.99 g/mol ≈ 0.0500 mol
Since NaOH dissociates into Na+ and OH- ions in water, the number of moles of NaOH is equal to the number of moles of OH- ions: 0.0500 mol.
In neutral water, the concentration of hydronium ions is equal to the concentration of hydroxide ions. Therefore, [H3O+] = [OH-].
We can use the formula for molarity (M) to calculate the concentration of hydronium ions, [H3O+], in the solution:
Molarity (M) = moles of solute / volume of solution in liters
The moles of OH- ions in 2.00 L of solution is:
moles of OH- = (0.0500 mol / 2.00 L) = 0.0250 M
Therefore, the concentration of hydronium ions, [H3O+], in this solution is 0.0250 M.
To determine the concentration of hydronium ions ([H3O+]) in a solution, we need to know the molarity of the solution.
The molarity (M) is defined as the number of moles of solute dissolved in one liter of solution. We are given that 2.00 g of NaOH is dissolved in 2.00 L of solution. However, NaOH is a strong base and dissociates completely in water, so we can consider it as a source of hydroxide ions (OH-).
The balanced equation for the dissociation of NaOH in water is:
NaOH → Na+ + OH-
Since NaOH dissociates completely, the concentration of hydroxide ions ([OH-]) in the solution will be the same as the molarity of NaOH. To find the molarity, we need to convert the mass of NaOH into moles.
The molar mass of NaOH is calculated as follows:
Na (22.99 g/mol) + O (16.00 g/mol) + H (1.01 g/mol) = 39.99 g/mol
Converting the mass of NaOH to moles:
2.00 g NaOH * (1 mol NaOH / 39.99 g NaOH) = 0.0500 moles NaOH
Since we have 0.0500 moles of NaOH dissolved in 2.00 L of solution, the molarity (M) is calculated by dividing moles by liters:
Molarity (M) = moles of solute / volume of solution in liters
M = 0.0500 moles NaOH / 2.00 L = 0.0250 M NaOH
Since the NaOH completely dissociates in water, the concentration of hydroxide ions ([OH-]) is also 0.0250 M.
Now, to determine the concentration of hydronium ions ([H3O+]) in the solution, we can use the fact that water autoprotolyses to some extent. This autoprotolysis is represented by the equation:
2H2O ⇌ H3O+ + OH-
In a neutral solution, the concentration of hydronium ions ([H3O+]) is equal to the concentration of hydroxide ions ([OH-]). Therefore, the concentration of hydronium ions in this NaOH solution is 0.0250 M.
Forgot a step:
pH=-log[H3O+], so
10^-pH=H3O+
NaOH is a strong base, so it will completely dissociate into OH-.
pOH+pH=14, so 14-pOH=pH
Solve for pOH
2.00g of NaOH*(1 mole of NaOH/39.997 g)= moles of NaOH
moles of NaOH/2.00L=molarity of NaOH
-log[molarity of NaOH]=pOH
14-pOH=pH