. What would be the molarity of the solution produced when 276 mL of 6.00 M NaOH is combined with 125 mL of a 2.00 M NaOH solution?
M final solution = total mols/total liters.
mols soln 1 = M x L = ?
mols soln 2 = M x L = ?
total mols = ?
volume soln 1 = 0.276 L
volume soln 2 = 0.125 L.
total volume = ?
4.75M NaOH
To find the molarity of the resulting solution, you need to use the formula for dilution:
M₁V₁ = M₂V₂
where:
M₁ = initial molarity of the solution
V₁ = initial volume of the solution
M₂ = final molarity of the solution
V₂ = final volume of the solution
In this case, the initial solution is 6.00 M NaOH with a volume of 276 mL, and the second solution is 2.00 M NaOH with a volume of 125 mL. The final volume will be the sum of the initial volumes, which is:
V₂ = V₁ + V₂
V₂ = 276 mL + 125 mL
V₂ = 401 mL
Next, we can plug in the values into the dilution formula:
M₁V₁ = M₂V₂
(6.00 M)(276 mL) = M₂(401 mL)
Now, solve for M₂:
M₂ = (6.00 M)(276 mL) / (401 mL)
M₂ ≈ 4.12 M
Therefore, the molarity of the resulting solution when 276 mL of 6.00 M NaOH is combined with 125 mL of a 2.00 M NaOH solution is approximately 4.12 M.