Average percent of KHP Standard deviation_ Calculations of percent KHP and stand

Question

Average percent of KHP Standard deviation_ Calculations of percent KHP and standard deviation (show using equations with units) QUESTIONS 1. Write the balanced chemical equation for the reaction of KHP with NaOH 2. Suppose your laboratory instructor inadvertently gave you a sample of KHP contaminated with NaCI to use in standardizing your NaOH. How would this affect the molarity you calculated for your NaOH solu- tion? Justify your answer. 3. How many grams of NaOH are needed to prepare 500 mL of 0.125 MNaOH? 4. A solution of malonic acid, H,C3H204, was standardized by titration with 0.1000 M NaOH solution. If 20.76 mL of the NaOH solution is required to neutralize completely 12.95 mL of the malonic acid solu- tion, what is the molarity of the malonic acid solution? 5. Sodium carbonate is a reagent that may be used to standardize acids in the same way you used KHP in this experiment. In such a standardization, it was found that a 0.512 g sample of sodium carbonate re- quired 26.30 mL of a sulfuric acid solution to reach the end point for the reaction. Na CO,(ag)+H,SO4(ag)H,00)+CO2(g)+Na,SO4(aq) Copyright © 2015 Pearson Education, Inc.

Explanation / Answer

Answer 1:

Potassium hydrogen pthalate (KHP) is an acid. Sodium hydroxide (NaOH) is a base. An acid and a base reacts to form salt and water.

The formula for KHP is =  KHC8H4O4

The balanced equation of the reaction can be written as:

NaOH(aq) + KHC8H4O4(aq) = KNaC8H4O4(aq) + H2O(l)  

Answer 2:

If KHP is contaminated with NaCl, then some NaCl (say x moles) would react with KHP according to the following equation:

NaCl + KHC8H4O4 = Na + ClH + KC8H4O4

So, you are actually weighing less amount of KHP as it has already reacted with NaCl. Hence, you would estimate a lower molarity of NaOH as there are less H+ ions to react with OH- of NaOH ( see answer 1).

Answer 3:

Molarity = Mass of the compound in grams/ Molar mass of the compound

Molar mass of NaOH = 40 g

This means, for 1 M solution of NaOH, 40 g NaOH has to be dissolved in 1 l of water.

Therefore, for 0.125 M solution of NaOH, amount of NaOH required = 40 * 0.125 g = 5 g

Now, 5 g NaOH is required in 1000 ml water.

Hence, for 500 ml, amount of NaOH required is = 1000*5 / 500 = 2.5 g of NaOH

Answer 4:

H2C3H2O4 + 2NaOH = Na2C3H2O4 + 2H2O

From the reaction, it can be seen that one mole of malonic acid is completely neutralized by 2 moles of NaOH.

Molar mass of NaOH = 40 g/ mole

Molar mass of malonic acid = 104 g/ mole

So, it can be said that 104 g of malonic acid is neutralized by (2*40) 80 g of NaOH.

Hence, we find out grams of NaOH consumed .

20.76 ml of 0.1 M NaOH contains 0.1 *40*20.76/ 1000 g of NaOH

= 0.08304 g

80 g NaOH neutralizes 104 g Malonic acid

So, 0.08304 g NaOH neutralizes 104* 0.08304/80 = 0.108 g Malonic acid

Therefore molarity of malonic acid = 0.108*12.95/104 = 0.013 M

Answers 5 and 6: Cannot be answered in this question as this question has more than 4 subparts. Please ask questions 5 and 6 as a separate question.

Thank you!

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