Pre-Laboratory Assignment source for a 1. Read TECH Measurement this series or a

Question

Pre-Laboratory Assignment source for a 1. Read TECH Measurement this series or another authoritative discussion 325, volume of Liquids, in of titration and the use of a pipet and a buret. 2. What are two special precautions for storing sodium h solutions? 3. The exact concentration of a dilute NaoH solution is determined by reaction with another substance. dilute can't we just dilute the concentrated reagent and calculate the concentration of the carefully solution? Why must standard HCl be put in a dry beaker? 5. a) What is the name of the indicator used for this titration? What color is the indicator in a hydrochloric acid solution?

Explanation / Answer

1) The special precuations for storing NaOH solutions is it must be stored inplastic botles , otherwise the glass of glass bottle dissolves slowly in NaOH. That is why usually the lids of NaOH containers get stuck.

secondly it absorbs CO2 from the atmosphere and changes to na2CO3 , thus its concnetraion changes.

2)The exact cncentration of NaOH solution has to be calulated by using a standard solution of HCl ,due ot

the fact that it is not primary standard, its concentration changes on storing , either by reaction with glass or due to absorption of CO2 from atmosphere.

3) We need to put the standard HCl in a dry flask only, as a few drops of water adhered to wet flask can change the concentration of standard Hcl and thus error in values.

4) The indicator used for HCl vs. NaOH titration is either methyl orange or phenolphthalein.

5) Methyl orange indicator islight/pale pink in acid solution and phenolphthalein is colorless in acid solution.

6) If methyl orange is the indicator and acid in burette, it should be yellow to pale pink.

If the indicator is phenolphthalein and acid in burette, it should be pink to colorless.

Q1)

a) volume of HCl in L = Vin mL x1L /1000mL

= 25.0mL x1 L/ 1000mL

= 0.025L

volume of NaOh = burette reading final - burette reading initial

= 23.58-0.06

= 23.52mL

no volume of NaOH in L = 23.52mL x 1L / 1000mL

= 0.02352mL

b)

mmoles of base = mmoles of acid for neutralization reactions(titrations)

23.52mL x M = 25.0 x 0.1010M

Hence M = 0.1073 M

c) The three other titrations gave the molarities of base as 0.1066, 0.1072 ,0.1077 M.

thus the average of four molarityvalues of NaOH is

=[ 0.1073 + 0.1066+ 0.1072 +0.1077] /4

=0.1072 M

D) If the initial burette reading was not recorded , then the volume of NaOH is measred as 23.58mL

Then molarity of NaOH = 25.0x 0.1010 /23.58

=0.1070 M

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