POST LAB QUESTIONS You prepare a diluted NaOH solution by adding 15 mL of 3 M Na

Question

POST LAB QUESTIONS You prepare a diluted NaOH solution by adding 15 mL of 3 M NaOH to 435 mL of distilled water and stir for several minutes the molarity. solu . In a clean dry beaker you obtain about 75 mL of standardized HCl and record You take two burets, rinse one with the standardized HCl and one with the diluted NaOH tion. Fill one buret with the HCl and the other with NaOH. 1. Prior to the titration the standard HCl solution (~-15.00 mL) was diluted to a total volume of 65.00 mL a. (6 pts) Calculate the molarity of HCI in the diluted solution. b. (6 pts) Calculate the normality of HCl in the diluted solution. 2, (2 pts) For the titration 5 drops of 0.2% phenolphthalein indicator are added to the beaker containing the 65 mL HCl solution from question I. What color is the HCl solution in the beaker? 3. (6 pts) The initial burette reading is 1.45 mL NaOH titrant. The volume of NaOH titrant added to reach the endpoint is 22.57 mL. Calculate the molarity of the NaOH solution.

Explanation / Answer

NaOH + HCl = NaCl + H2O

Step 1 Calculate the molarity of NaOH

(Mcon) (Vcon) = (Mdil) (Vdil)

(3 M) (15 mL) = (Mdil) (435 mL)

(Mdil) = (3 M) (15 mL) /(435 mL)

(Mdil) = 0.1034 M

Step 2 Calculate the number of moles of in 0.1034 M 435 mL NaOH by using formula n = C*V

Moles = 0.1034 x 0.435 = 0.04498

Step 3 Calculate molarity of 45 ml standardized HCl

Moles of NaOH = moles of HCl = 0.04498

Molarity of 45 ml standardized HCl = 0.04498moles/0.045L = 1.0 M

1a) Molarity of dilute HCl is calculated by using formula

(Mcon) (Vcon) = (Mdil) (Vdil)

(1.0 M HCl)(15mL HCl) = (Mdil) (65 mL)

(Mdil) = (1.0 M HCl)(15mL HCl) / (65 mL)

(Mdil) = 0.231 M

b)

Molarity of an acid can easily be converted to Normality by multiplying Molarity by the number of hydrogen ions in the acid

Number of hydrogen ion in HCl =1

N = (M)(number of hydrogen or hydroxide ions)

N = 0.231 x 1 = 0.231

Normality of HCl = 0.231 N

2)If 5 drops of 0.2% phenolphthalein is added to 0.231 M 65 ml HCl, the solution remains colorless.

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