A 5.00-mL sample of a sulfuric acid solution of unknown concentration is titrated with a 0.1401 M Sodium Hydroxide solution. A volume of 5.99 mL of the base was required to reach the endpoint. What is the concentration of the unknown acid solution? Write the neutralization reaction, balance it, name the products THEN Solve the problem
write the equaiton balance it, name the products.
I will be happy to check your calculations
m1v1= m2v2
5* 0.1401 = m2 *5.99
m2 =0.1169449081803005 M
H2SO4 + 2NaOH =Na2SO4+2H2O neutralisation reaction
To solve this problem, we need to follow a few steps.
Step 1: Write the balanced neutralization reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH):
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
Step 2: Determine the stoichiometry of the reaction. The balanced equation shows that 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide.
Step 3: Calculate the number of moles of sodium hydroxide used in the titration using the given volume and molarity:
moles of NaOH = volume (L) x concentration (M)
moles of NaOH = 0.00599 L x 0.1401 M
Step 4: Since the stoichiometric ratio between sulfuric acid and sodium hydroxide is 1:2, the number of moles of sulfuric acid is half the number of moles of sodium hydroxide:
moles of H2SO4 = 1/2 x moles of NaOH
Step 5: Calculate the concentration of sulfuric acid by dividing the moles of sulfuric acid by the initial volume of the acid solution:
concentration (M) = moles of H2SO4 / volume (L)
Given that the initial volume of the sulfuric acid solution is 0.00500 L, you can substitute the values into the equation to find the concentration of the unknown acid solution.
Note: Make sure to perform the arithmetic calculations to obtain the final answer.