A 15.0mL solution of H2SO3 is neutralized by 12.0mL of a standardized 1.00M NaOH solution. What is the concentration of the H2SO3 solution?
Please reply with the steps on how to do this question so I can learn. I can't find examples or anything related in my textbooks and I take it via correspondence, so no class instructor to ask for help. Thank you
H2SO3 + 2NaOH ==> Na2SO3 + 2H2O
mols NaOH = M x L = ?
Convert mols NaOH to mols H2SO3 so
mols H2SO3 = 1/2 mols NaOH (look at the coefficients).
Now M H2SO3 = moles H2SO3/L H2SO3. Solve for M H2SO3.
To find the concentration of the H2SO3 solution, you can use the concept of stoichiometry and the equation for the neutralization reaction between H2SO3 and NaOH.
Step 1: Write and balance the neutralization reaction equation:
H2SO3 + 2NaOH → Na2SO3 + 2H2O
Step 2: Calculate the number of moles of NaOH used in the neutralization reaction using its concentration and volume:
moles of NaOH = concentration of NaOH x volume of NaOH
= 1.00M x 0.0120L
= 0.0120 moles
Step 3: Since the stoichiometry of the reaction tells us that one mole of H2SO3 reacts with two moles of NaOH, we can determine the number of moles of H2SO3:
moles of H2SO3 = (moles of NaOH) / 2
= 0.0120 moles / 2
= 0.0060 moles
Step 4: Calculate the concentration of H2SO3 using the volume of the H2SO3 solution:
concentration of H2SO3 = (moles of H2SO3) / volume of H2SO3
= 0.0060 moles / 0.0150L
= 0.400 M (rounded to three significant figures)
Therefore, the concentration of the H2SO3 solution is 0.400 M.