You prepare a diluted NaOH solution by adding 15 mL of 3 M NaOH to 435 mL of dis

Question

You prepare a diluted NaOH solution by adding 15 mL of 3 M NaOH to 435 mL of distilled water anc stir for several minutes. In a clean dry beaker you obtain about 75 the molarity. Y solution. Fill one buret with the HCl and the other with NaOH. mL of standardized HCl and record ou take two burets, rinse one with the standardized HCl and one with the diluted 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 HCl in the diluted solution. b. (6 pts) Calculate the normality of HCI 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 HCI solution from question 1. What color is the HCI 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

1) Molarity is defined as the no.of moles of a substance dissolved in 1000mL of solution. Here the final volume of NaOH is given by 15 + 435 = 450mL. Thus, the final concentration of NaOH solution will be as per Law of equivalence as V1M1 = V2M2 and M2 = (V1M1)/V2 = (15 x 3)/450 = 0.1M.

a) It is given that the volume of 0.1M NaOH needed to reach endpoint with HCl is 22.57 - 1.45mL = 21.12mL. The law of equivalence dictates that the concentration of a certain volume of one substance reacting completely with certain volume and concentration of another is related as VaMa = VbMb and Ma = (VbMb)/Va with a representing acid and b for base. This gives the concentration of HCl, Ma = (VbMb)/Va = (21.12 x 0.1)/65 = 0.0325M.

b) Normality of acids and bases are found as the (molarity/equivalents) with the equivalents being equal to the proticity of acids and basicity of bases. HCl being a monoprotic acid has molarity equal to molarity as 1 eq equals one normal. Thus, the normality of HCl = Molarity/1 = 0.0325/1 = 0.0325N after dilution.

2) Phenolphthalein is an acid-base indicator which is a weak acid with ionizable functional groups. In acidic medium, these functional groups remain undissociated and that form of the indicator is colourless. In basic medium, the groups are ionized and the salt of the indicator forms, which is pink in colour. Since the beaker contains HCl, an acid, no ionization of phenolphthalein takes place and so the solution in the beaker will be colourless.

3) In the first question, it was calculated that the molarity of NaOH solution after dilution was 0.1M using the law of equivalence. Though this parameter is asked now, without calculating this, the first question cannot be answered.

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