In the titration of a 0.101 M sodium hydroxide solution, 19.02 mL of 0.130 M sulfuric acid solution was required to neutralize the sodium hydroxide in reactions that replace both hydrogen ions of the sulfuric acid. Calculate the number of milliliters of sodium hydroxide solution needed in the reaction.
Thank you sooo much. I really appreciate the help everyone gives on here:)
Write the equation and balance it.
2NaOH + H2SO4 ==> Na2SO4 + 2H2O
moles H2SO4 = M x L = ??
moles NaOH = 1/2 that (from the coefficients in the balanced equation).
MNaOH = moles NaOH/LNaOH.
Solve for L NaOH and convert to mL.
0.204 L
To solve this problem, we can use the concept of stoichiometry. Stoichiometry is a calculation of the quantities of reactants and products involved in a chemical reaction.
Here are the steps to solve the problem:
1. Write the balanced chemical equation for the reaction between sodium hydroxide (NaOH) and sulfuric acid (H2SO4):
2NaOH + H2SO4 -> Na2SO4 + 2H2O
2. Calculate the number of moles of sulfuric acid used:
Moles of H2SO4 = (0.130 M) x (19.02 mL) x (1 L / 1000 mL)
3. Use the stoichiometry of the balanced equation to find the number of moles of NaOH required to react with the given amount of sulfuric acid. From the balanced equation, we can see that the mole ratio between H2SO4 and NaOH is 1:2:
Moles of NaOH = (2 moles H2SO4) x (moles of H2SO4)
4. Finally, calculate the volume of sodium hydroxide solution needed:
Volume of NaOH solution (in mL) = (moles of NaOH) x (1000 mL / 0.101 M)
By plugging in the values and performing the calculations, you will obtain the number of milliliters of sodium hydroxide solution required in the reaction.