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Chapter 15: Problem 80

To standardize a hydrochloric acid solution, it was used as titrant with a solid sample of sodium hydrogen carbonate, NaHCO \(_{3} .\) The sample had a mass of \(0.3967 \mathrm{g},\) and 41.77 \(\mathrm{mL}\) of acid was required to reach the equivalence point. Calculate the concentration of the standard solution.

Short Answer

Expert verified
The concentration of the hydrochloric acid solution is 0.113 M.

Step by step solution

01

Calculate the number of moles of sodium bicarbonate

Use the molar mass of sodium bicarbonate (NaHCO3) to calculate the number of moles in the given sample. The molar mass of NaHCO3 is approximately 84 g/mol. So, moles of NaHCO3 = mass of NaHCO3 / molar mass of NaHCO3 = 0.3967 g / 84 g/mol = \(4.72 \times 10^{-3}\) mol.
02

Understanding the reaction stoichiometry

From the balanced reaction between hydrochloric acid (HCl) and sodium bicarbonate (NaHCO3), you can see that one mole of HCl reacts with one mole of NaHCO3. Therefore, moles of HCl used = moles of NaHCO3 = \(4.72 \times 10^{-3}\) mol.
03

Calculate the concentration of HCl

Again applying the molarity formula (Molarity = Moles of solute / Volume of solution in liters), we can calculate the concentration of HCl. Remember to convert the volume from mL to Liters (1 L = 1000 mL). Therefore, concentration of HCl = moles of HCl / volume of HCl in L = \(4.72 \times 10^{-3}\) mol / 41.77 mL \times 1L/1000mL = 0.113 M.

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