Question
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?
Answers
GPT 3.5
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.
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.