Hello, I need help calculating these solutions. For the lab we used 50.0 ml of .

Question

Hello, I need help calculating these solutions.

For the lab we used 50.0 ml of .100 M NaOH for both acids

so the volume of .100 M NaOH is 50 ml

The two acids below is unknown but i used 10 ml of both acids. I added 10.0 ml of acid in 50.0 ml diluted water. i pipetted the 10 ml acid into a 50 ml diluted water separate times not together cuz it was two different trails since I we had to find both ph

hydrochloric acid, HCl, solution, unknown molarity

acetic acid, HC2H3O2, solution, unknown molarity

DATA ANALYSIS Report your answers to three (3) significant figures Calculate the molar amounts of NaOH used in the reaction with the HCl solution and with the HC2H302 solution. (record your answer in Table 3) 1. NaOH and HCl NaOH and Acetic Acid 2. Calculate the molar concentration (molarity) of the HCl solution and the HC2H302 solution. (record your answer in Table 3) NaOH and HCI NaOH and Acetic Acid 3. Compare the actual molar concentrations of your two acid solutions with your calculated molarities. Were the calculated molarities of your acid solutions within a reasonable range (about 5%) of the actual values? If not, suggest reasons for the inaccuracy.

Explanation / Answer

1) We added 50.0 mL of 0.100 M NaOH solution to both the acids; hence, the molar amounts of NaOH used in both the titrations = (50.0 mL)*(1 L/1000 mL)*(0.100 M) = 0.005 mole (ans).

2) Consider the reactions of NaOH with HCl and HC2H3O2 as below.

HCl (aq) + NaOH (aq) --------> NaCl (aq) + H2O (l)

HC2H3O2 (aq) + NaOH (aq) -------> NaC2H3O2 (aq) + H2O (l)

As per the stoichiometric equations,

1 mole NaOH = 1 mole HCl = 1 mole HC2H3O2.

Therefore,

0.005 mole NaOH = 0.005 mole HCl = 0.005 mole HC2H3O2

10.0 mL of both the acids were taken; hence molarity of HCl = molarity of HC2H3O2 = (0.005 mole)/[(10.0 mL)*(1 L/1000 mL)] = 0.500 M (ans).

3) Need data to answer this question.

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