If it takes 15.0 mL of 0.40 M NaOH to neutralize 5.0 mL of sulfurous acid, what is the molarity of the acid solution?
To find the molarity of the acid solution, we can use the formula:
Molarity = (moles of solute) / (volume of solution in liters)
First, let's find the moles of NaOH used in the reaction:
moles of NaOH = Molarity NaOH x Volume NaOH (in liters)
moles of NaOH = 0.40 mol/L x (15.0 mL / 1000 mL/L)
moles of NaOH = 0.40 x 0.015
moles of NaOH = 0.006 mol
Since NaOH and the acid react in a 1:1 ratio, the moles of sulfurous acid is also 0.006 mol.
Now, let's find the molarity of the acid solution:
Molarity acid = moles of acid / Volume of acid solution (in liters)
Molarity acid = 0.006 mol / (5.0 mL / 1000 mL/L)
Molarity acid = 0.006 / 0.005
Molarity acid = 1.2 M
Therefore, the molarity of the acid solution is 1.2 M.