Show ALL work What is a secondary standard? Describe an acid-base indicator. How

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What is a secondary standard? Describe an acid-base indicator. How does it work? Balance the equation: NaOH + H2C2O4 rightarrow H2O + Na2C2O4 In a series of titration experiments, why is it that you always work with the smallest sample first? Explain. How many grams of H2C2O4 are needed to completely neutralize 30 mL of a 0.100 M NaOH solution? You want to prepare a 500 mL 0.20 M NaOH solution and your only source is a 6.0 M NaOH solution. How will you prepare the solution? Show calculations; explain the process. With the following data calculate the molarity of the NaOH solution:

Explanation / Answer

3. A secondary standard is a standard that is prepared in the laboratory for a specific analysis. Its concentration used to change, not exacltly to our calculation. It is usually standardized against a primary standard.

4. An acid-base indicator is a weak acid or a weak base. The undissociated form of the indicator is a different color than the iogenic form of the indicator. An Indicator does not change color from pure acid to pure alkaline at specific hydrogen ion concentration, but rather, color change occurs over a range of hydrogen ion concentrations. This range is termed the color change interval. It is expressed as a pH range.

Weak acids are titrated in the presence of indicators which change under slightly alkaline conditions. Weak bases should be titrated in the presence of indicators which change under slightly acidic conditions.

5. 2NaOH + H2C2O4 --------------------> 2H2O + 2Na2C2O4

6. Smallest sample used to use while doing a series of titration due to less probable of loosing sample.

7. From the ballance euqation , 2 mol of NaOH is requred by 1 mol of H2C2O4 for neutralization.

So, H2C2O4 concentration will be 0.1/2 = 0..05 mol/L

Then, amount of H2C2O4 (w) will be 0.1 = (w x 1000) / (90 x 1000) = 9 g [where molar mass of H2C2O4 = 90g/mol]

8. Using this equation

V1S1 = V2S2

Where,

V1 = Initial volume = ?

S1 = Initial strength = 6.0 M

V2 = Diluted volume = 500 mL

S2 = Diluted strength = 0.2 M

So, V1 x 6 = 500 x 0.2

=> V1 = 16.67 mL

9. Mass of wt bottle = 35.5476 g

Mass of wt bottle - mass of H2C2O4 = 37.3114 g

So, mass of H2C2O4 will be = 37.3114 -37.5476 = 0.2346 g

Final reading of burett NaOH = 34.75 mL

Initial reading of burett NaOH = 0.24 mL

Volume of NaOH used = 34.75 - 0.24 =34.51 mL

Concentration of NaOH (S2) will be (using V1S1 =V2S2 formula)

V1 x 0.2346/ (90 x V1) = 34.51 x S2

So, S2 = 7.56 x 10^-5 M

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