i need to finished this but dont understand how to solve for them & i havr to co

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i need to finished this but dont understand how to solve for them & i havr to complete this by today

REPORT PAGE EXPERIMENT 13-STANDARDIZATION OF A BASIC SOLUTION CHEM 1405 SECTION: 10O Staple this page behind the pages for Experiment 13 from y Sample #2 sa 1 Weight of KHP sample asea O.ST 2 Molecular Weight (MW) of g/mol KHP (formula KCH,O4) 3 Final buret reading 4 Initial buret reading 5 Volume of NaOH solution 35.5 m. 30.5 3 3s nL mL mL' used (in milliliters) 6 Volume of NaOH solution used (in liters) 7 Molarity of NaOH solution 8 Average molarity (average of closest two) All readings in ml should be recorded to two decimal places (x.xx mL). Molarity of NaOH solution: M Rac, acid X Vbase www.templejc.edu

Explanation / Answer

Ans. # Balanced Reaction:   KHC8H4O4(aq) + NaOH(aq) -------> KNaC8H4O4(aq) + H2O(l)

According to the stoichiometry of balanced reaction, 1 mol NaOH neutralizes 1 mol KHP.

#. Sample 1: Moles of KHP taken = Mass / Molar mass

                                                            = 0.601 g / (204.22 g/mol)

                                                            = 0.002943 mol

# Volume of NaOH consumed = 35.5 mL = 0.0355 L

# Following stoichiometry of neutralization, at titration endpoint, the moles of KHP must be equal to that of NaOH in the volume of NaOH consumed to reach the endpoint.

So, moles of NaOH in 35.5 mL NaOH solution = 0.002943 mol

Now,

            Molarity of NaOH = Moles of NaOH / Volume of solution in liters

                                                = 0.002943 mol / 0.0355 L

                                                = 0.0829 mol/ L

                                                = 0.0829 M

#. Sample 2: Moles of KHP taken = 0.587 g / (204.22 g/mol) = 0.002874

# Volume of NaOH consumed = 32.5 mL = 0.0325 L

# Following stoichiometry of neutralization, at titration endpoint, the moles of KHP must be equal to that of NaOH in the volume of NaOH consumed to reach the endpoint.

So, moles of NaOH in 32.5 mL NaOH solution = 0.002874 mol

Now,

            Molarity of NaOH = 0.002874 mol / 0.0325 L = 0.0884 M

                                               

#. Sample 3: Moles of KHP taken = 0.599 g / (204.22 g/mol) = 0.002933

# Volume of NaOH consumed = 37.35 mL = 0.03735 L

# Following stoichiometry of neutralization, at titration endpoint, the moles of KHP must be equal to that of NaOH in the volume of NaOH consumed to reach the endpoint.

So, moles of NaOH in 37.35 mL NaOH solution = 0.002933 mol

Now,

            Molarity of NaOH = 0.002933 mol / 0.03735 L = 0.0785 M

## Average molarity of NaOH = (0.0829 + 0.0884 + 0.0785) M / 3 = 0.0833 M

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