QUESTION4 2 points Save Answer The four chemistry friends Kemmi, Doc, Stu, and N

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QUESTION4 2 points Save Answer The four chemistry friends Kemmi, Doc, Stu, and Noke perform a standardization of sodium hydroxide solution using potassium hydrogen phthalate (KHP) and the procedure described in Lab 1 of this Module. Use Doc Inmaking's data to help him perform each step of the calculations over the next six questions Doc weighs 2.9842 g of potassium hydrogen phthalate solid for his titration. Calculate the moles of potassium hydrogen phthalate he used. (Answer to 5 significant figures, units of mol. Hint: the molar mass of potassium hydrogen phthalate is given in the Introductory Material of the Module.) QUESTION 5 2 points Saved During the titration the potassium hydrogen phthalate undergoes an acid-base reaction with the sodium hydroxide. What is the mole ratio of potassium hydrogen phthalate to sodium hydroxide in the reaction, according to the chemical equation? O 1 mole KHP: 1 mole NaOH O l mole KHP : 2 mole NaOH O 3 mole KHP: 1 mole NaOH 3mole KHP : 2 mole NaOH O 2 mole KHP : 1 mole NaOH O 2 mole KHP: 3 mole NaOH 2 points Save Answer QUESTION 6 answers to calculate the number of moles of sodium hydroxide reacting with Doe's KHP sample. (Answer to 5 significant figures, Use the two units of mol.) 2 points Save Answer QUESTION 7

Explanation / Answer

QUESTION 4: We can calculate moles of KHP as follows:

moles of KHP = mass (g) / molar mass KHP = 2.9842g/204.22 g/mol = 0.014613 moles of KHP.

QUESTION 5: KHP/NaOH mole ratio is 1:1

The reaction is : KHP + NaOH ----> KNaP + H2O

One mole of KHP reacts with one mole of NaOH.

Then : 1 mole KHP: ! mole NaOH

QUESTION 6: According to stoichiometry:

moles of KHP = moles of NaOH

Then , moles of NaOH = 0.014613 moles NaOH

QUESTION 7: Initial reading= 2.64mL and final reading = 23.58mL

Volume of NaOH = final reading - initial reading = 23.58mL - 2.64mL = 20.94mL.

QUESTION 8:Molarity of NaOH can be calculated as follows:

molarity of NaOH = moles / Volume in L = 0.014613moles/20.94 x 10-3 L

molarity of NaOH = 0.6979mol/L .

QUESTION 9: Value 0.6034 mol/L should be omitted. Then we can calculate the average as follows:

Average concentration of NaOH = (0.7014 + 0.6824 + 0.6979 ) mol/L / 3

Average concentration of NaOH = 0.6939 mol/L NaOH.

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