A student titrated an unknown hydrochloric acid solution with a 0.115 M sodium hydroxide solution. They measured 8.00 mL of acid and it took 9.68 mL of the sodium hydroxide solution to reach the endpoint. Using the balanced equation that moles of acid equals moles of base, and the formula relating molarity, moles, and volume, the molarity of the hydrochloric acid solution was calculated to be 0.139 M.

1. 2Calculate the molarity (M) of an unknown solution of hydrochloric acid (HCI). This experiment
is similar to the one done in the lab session., The acid solution was titrated in a flask with a
solution of NaOH added from the buret until the end point (light pink color in the flask) was seen
. Followings are the data measured by a student. (3 points) Molarity (M) of NaOH (given)0.115
MB Volume of acid ( measured) in the flask 8.00 mL Volume of NaOH required to neutralize
the acid ( Final-Initial)-- 9.68 mL Calculate the molarity (M A) of the acid. Show your work with
formula and units. Equation and Rearranged. Eq Calculation Steps with units
Solution
In this titration reaction,
Moles of NaOH = Moles of HCl
[Moles = Molarity*Volume]
Hence, Molarity of NaOH*Volume of NaOH = Molarity of HCl*Volume of HCl
0.115 M*9.68 mL = Molarity of HCl*8.00 mL
Molarity of HCl = 0.139 M