6.00 g of NaOH are dissolved in water to make 4.00 L of solution. What is the concentration of hydronium ions, [H3O+], in this solution?

To determine the concentration of hydronium ions ([H3O+]) in the solution, you need to consider the dissociation of sodium hydroxide (NaOH) in water.

The balanced chemical equation for the dissociation of NaOH is:
NaOH → Na+ + OH-

Notice that for every one molecule of NaOH that dissociates, it forms one hydronium ion (H3O+) and one hydroxide ion (OH-).

Given that 6.00 g of NaOH is dissolved in 4.00 L of water, we need to convert the mass of NaOH to moles, and then determine the concentration of hydronium ions.

Step 1: Convert the mass of NaOH to moles.
The molar mass of NaOH is:
Na: 22.99 g/mol
O: 16.00 g/mol
H: 1.01 g/mol
So, the total molar mass of NaOH is 22.99 + 16.00 + 1.01 = 40.00 g/mol.

To convert the mass to moles, divide the given mass by the molar mass:
Number of moles of NaOH = Mass of NaOH / Molar mass of NaOH
Number of moles of NaOH = 6.00 g / 40.00 g/mol

Step 2: Calculate the molarity (concentration) of NaOH.
Molarity (M) = Number of moles / Volume of solution (in liters)

In this case, the volume of the solution is given as 4.00 L.

Molarity (M) = (6.00 g / 40.00 g/mol) / 4.00 L

Step 3: Calculate the concentration of hydronium ions ([H3O+]).
Since NaOH dissociates in a 1:1 ratio to produce hydronium ions, the concentration of hydronium ions is equal to the molarity of NaOH.

Therefore, the concentration of hydronium ions is equal to the molarity of NaOH, which is:
[H3O+] = (6.00 g / 40.00 g/mol) / 4.00 L

You can now substitute the values into the equation and calculate the concentration of hydronium ions.

To find the concentration of hydronium ions, [H3O+], in the solution, we need to calculate the molarity (M) of the solution.

The molarity of a solution is calculated using the formula:

Molarity (M) = moles of solute / volume of solution (in liters)

First, let's calculate the moles of NaOH:

Moles = mass (g) / molar mass (g/mol)

The molar mass of NaOH is:
Na = 22.99 g/mol
O = 16.00 g/mol
H = 1.01 g/mol

Molar mass of NaOH = 22.99 + 16.00 + 1.01 = 40.00 g/mol

Moles of NaOH = 6.00 g / 40.00 g/mol = 0.15 mol

Now, let's calculate the molarity:

Molarity (M) = moles of solute / volume of solution (in liters)

Molarity (M) = 0.15 mol / 4.00 L = 0.0375 M

Therefore, the concentration of hydronium ions, [H3O+], in the solution is 0.0375 M.

mols NaOH = grams/molar mass = ?

Then M = molarity = mols/L solution. You know mols and L, solve for M. That will be the OH^- since NaOH is a strong electrolyte.
Convert to H^+ (H3O^+) with
(H^+)(OH^-) = Kw = 1E-14
You know Kw and OH, solve for H..