Question

PART A. Standardization of NaOH(aq) Calculate the molar mass of KHP from the molecular formula: __204.22 g/mol______________

	Trial	Run 1	Run 2
Mass of KHP	0.326g	0.307g	0.314g
Calculate Moles of KHP
Calculate Moles of NaOH needed for the reaction
Initial volume of NaOH(aq) in mL	11.30ml	22.20ml	25.20ml
Final volume of NaOH(aq) in mL	21.60ml	31.60ml	35.00ml
Calculate the total volume of NaOH titrated in mL
Convert the total volume of NaOH titrated into L	xxxx
Calculate the concentration of NaOH in moles/L	xxxx

Calculate the average molarity of NaOH (in mol/L): _____________

PART B. Acetic Acid Concentration in Vinegar:

	Trial	Run 1	Run 2
Volume of vinegar in mL	2.00ml	2.00ml	2.00ml
Convert the volume of vinegar into L	xxxxx
Initial volume of NaOH(aq) in mL	12.50ml	39.50ml	35.60ml
Final volume of NaOH(aq) in mL	22.60ml	49.80ml	44.80ml
Calculate the total volume of NaOH titrated in mL
Convert the total volume of NaOH titrated into L	xxxxx
Calculate the moles of NaOH (use the average M from part A)	xxxxx
Calculate the concentration of vinegar in moles/L	xxxxx

Calculate the average molarity of vinegar (in mol/L): _______________

Answer 1

Solution:- Part A) Run one: moles are calculated on dividing the moles by molar mass.

Run two:

Run three:  

KHP and NaOH react in 1:1 mol ratio which is clear from the below balanced equation:

Since the mol ratio is 1:1, moles of NaOH will also be equivalent to the moles of KHP.

Run one:

Run two:

Run three:

Total Volume of NaOH used is the difference of final and initial volume readings.

Run one:- 21.60 mL - 11.30 mL = 10.30 mL

Run two: 31:60 mL - 22:20 mL = 9.40 mL

Run three:- 35.00 mL - 25.20 mL = 9.80 mL

Since, 1000 mL = 1 L

We divide the above volumes by 1000 to convert to L.

Run one:

Run two:

Run three:

To calculate the moles by Liters to get the molarity.

Run one:

Run two:  

Run three:

Average molarity of NaOH =

Part B) For each rune, 2.00 mL of vinegar is used. To convert it to L, we will divide by 1000 same as we did above for NaOH.

so, 2.00/1000 = 0.00200 L

Hence, the volume will be 0.00200 L for each run.

NaOH volume is calculated by taking the difference of final and initial volumes same as we did above for part A.

run one: 22.60 mL - 12.50 mL = 10.1 mL = 0.0101 L

Run two: 49.80 mL - 39.50 mL = 10.3 mL = 0.0103 L

Run three: 44.80 mL - 35.60 mL = 9.20 mL = 0.00920 L

moles of NaOH used are calculated on multiplying the average molarity from part A and the volume in L.

Run one:

Run two:

Run three:

NaOH and vinegar react in 1:1 mol ratio as shown in the below equation:

Since the mole ratio is 1:1, moles of vinegar will be equivalent to the moles of NaOH. and we divide the moles by Liters to get the molarity.

Run one:  

Run two:

Run three:

Average molarity of vinegar =