Standardization of NaOH with KHP: Write a balanced equation for the chemical rea

Question

Standardization of NaOH with KHP: Write a balanced equation for the chemical reaction(s) in the standardization of NaOH. Prepare a table: report the masses of standard used, the titration volumes and calculated NaOH molarity. Show sample calculations for finding the NaOH molarity. Below the table, report the average molarity of the NaOH, it's standard deviation, and the % relative standard deviation (%rsd). Titration of ASA with NaOH: Write a balanced equation for the chemical reaction(s) for the titration of ASA with NaOH. Prepare a table, including the following data for each titration (include the data for the titration you did with the pH meter): mass of tablet(s), titration volumes, blank volume, net titration volume, net moles of titrant, mass of ASA titrated, % wt/wt ASA, and the mass of ASA per tablet. Show sample calculations for determining the % wt/wt of ASA per tablet and the mass of ASA/tablet from net titration volume. Below the table, report the average %wt/wt ASA and its standard deviation, and the average ASA/tablet and its standard deviation.

Explanation / Answer

1.

(a)

Basi equation for the reaction in standardization of NaOH with KHP is:
NaOH + KHP ---> NaKP + H2O

Thus, reactants react in a 1:1 molar ratio

(b)

In this procedure, KHP is taken in a flask and a few drops of phenolpthalein indicator are added.

Assuming that you take 'x' mL of 'y' M KHP solution in flask

Here 'x' and 'y' are known beforehand.

Now. NaOH is added dropwise from the burette into the flask until a pink color appears. This is the equivalence point. At this point, record the volume of NaOH used uptil now. Call it 'z' mL.

If M denotes the molarity of the NaOH solution, then using the following equation allows us to solve for M:

x*y = M*z

Repeat this procedure several times to get many values for M

Then take the mean of those values and calculate the standard deviatio using the standard formula.

2.

Everything is same as in part 1, except that the reaction that is taking place is:

HC9H7O4 + NaOH ----> NaC9H7O4 + H2O

Here, HC9H7O4 represents ASA.

Here also the reactants react in a 1:1 molar ratio.

The rest of the procedure is same as above.

Revert back if you have any queries.

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