1) You are given solutions of HCl and NaOH and must determine their concentrations. You use 37.0mL of NaOH to titrate 100mL of HCl and 13.6 mL of NaOH to titrate 50.0mL of 0.0782 M H2SO4. Find the unknown concentrations.
Molarity of NaOH and molarity of HCL?
2) Sodium hydroxide is used extensively in acid-base titrations because it is a strong, inexpensive base. A sodium hydroxide solution was standardized by titrating 28.00mL of 0.1598 M standard HCl. The initial buret reading of the sodium hydroxide was 2.54mL, and the final reading was 39.85mL. What is the molarity of the base solution?
To find the unknown concentrations in both questions, we need to use the concept of stoichiometry and the equation of the chemical reaction that takes place during the titration.
1) Let's start with the first question. The balanced chemical equation for the reaction between HCl and NaOH is:
HCl + NaOH -> NaCl + H2O
Using the given information, we can set up an equation based on the stoichiometry of the reaction and the volume of solutions used:
(Molarity of HCl) * (Volume of HCl) = (Molarity of NaOH) * (Volume of NaOH)
For the first titration, we have:
(Molarity of HCl) * (100 mL) = (Molarity of NaOH) * (37.0 mL)
For the second titration, we have:
(Molarity of HCl) * (50.0 mL) = (Molarity of NaOH) * (13.6 mL)
Now, we can solve these two equations simultaneously to find the unknown concentrations.
2) For the second question, we again need to use the stoichiometry of the reaction and the volume of solutions used. The balanced equation for the reaction between HCl and NaOH is the same as before:
HCl + NaOH -> NaCl + H2O
Here, we are given the initial and final buret reading of the sodium hydroxide solution, which allows us to determine the volume of NaOH used in the titration:
Volume of NaOH = Final buret reading - Initial buret reading = 39.85 mL - 2.54 mL = 37.31 mL
Using this information, we can set up an equation:
(Molarity of HCl) * (Volume of HCl) = (Molarity of NaOH) * (Volume of NaOH)
Plugging in the known values, we have:
(0.1598 M) * (28.00 mL) = (Molarity of NaOH) * (37.31 mL)
Now, we can solve the equation to find the molarity of the NaOH solution.