Question

culate the molarity (M) of an unknown solution of hydrochlorie acid (HCT). This experiment similar to the one as 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) Volume of acid ( measured) in the flask Volume of NaOH required to neutralize the acid ( Final-Initial)18.10 ml. 0.215 Ma 8.00 mL Calculate the molarity (M a) of the acid. Show your work with formula and units. Equation and Rearranged. Eq Calculation Steps with units 3 Balance the following equations by adding coefficients before the formula of reactants and products. Classify the type of reaction. These reactions are similar to what were done in the lab. Write clearly. (9 points) Type of Reaction P12 (s) NaNO, (a) Pb(NO,)2 (aq) + Nal (aq + → yellow ppt b) Al (s) + CuSO4 (aq)Al(SO4)s (ag)+ Cu (s) reddish ppt. white strip high heat CaO (s) white solid forms c) Ca (s) +02(g) (foil)

Answer 1

2. The reaction associated with this experiment is, HCI + ŅaOH →NaCl + H2O Now, we will calculate the molarity of HCl by using the molarity equation shown below, (HCI) (NaOH) Now, we will substitute the values of molarity and volumes in the above equation. 0.215 Mx18.10 mI 8mL 3.89 M mL 8 mL M, 0.486 M Therefore, molarity of acid is 0.486 M.