Report for Exp 12 Name Section InstructorDate Titration Part I: Molarity of NaOH
ID: 635720 • Letter: R
Question
Report for Exp 12 Name Section InstructorDate Titration Part I: Molarity of NaOH solution Data Table Sample 1 Sample 2 Sample 3 5lIf needed Mass of the flask Mass of KHP Final buret reading Initial buret reading Volume of base used-8.nu^??.01 w~?| 26. 30 mg Calculations: Show calculations with units for Sample 1 only in the left column. Use the correct number of significant figures in the answers you enter in the other two (or three) columns. Show Sample 1 calculations below the bold print in each block below. Moles of KHP (molar mass-204.2) Sample1Sample 2 Sample 3 (if needed) Moles of the base required Molarity of NaOH Average molarity of baseExplanation / Answer
Sample 1
Sample 2
Sample 3
Moles of KHP (Molar mass = 204.2)
Moles = (grams of KHP)/(gram molar mass of KHP)
= (0.56 g)/(204.2 g/mol) = 2.7424*10-3 mole
2.7424*10-3 mole
2.6934*10-3 mole
2.6689*10-3 mole
Moles of the base required
The balanced chemical equation for the reaction is
NaOH (aq) + KHP (aq) --------> NaKP (aq) + H2O (l)
As per the stoichiometric equation,
Moles KHP = moles NaOH.
2.7424*10-3 mole
2.6934*10-3 mole
2.6689*10-3 mole
Molarity of NaOH
Molarity of NaOH = (moles of NaOH)/(volume of the base used in L)
= (27424*10-3 mole)/[(28.02 mL)*(1 L/1000 mL)] = 0.09787 mole/L ? 0.09789 M
0.09789 M
0.09972 M
0.10148 M
Average molarity of NaOH = 1/3*(0.09789 + 0.09972 + 0.10148) M = 0.09969 M ? 0.100 M (ans).
Sample 1
Sample 2
Sample 3
Moles of NaOH
Moles of NaOH = (volume of NaOH used in L)*(average molarity of NaOH)
= (33.05 mL)*(1 L/1000 mL)*(0.100 M) = 3.305*10-3 mole
3.305*10-3 mole
3.390*10-3 mole
3.392*10-3 mole
Moles of the acetic acid required
The balanced chemical equation for the reaction between acetic acid (HOAc) and NaOH is given below.
HOAc (aq) + NaOH (aq) ----------> NaOAc (aq) + H2O (l)
As per the stoichiometric equation,
Moles NaOH = moles HOAc
3.305*10-3 mole
3.390*10-3 mole
3.392*10-3 mole
Molarity of acetic acid in vinegar
Molarity of acetic acid = (moles of acetic acid)/(volume of vinegar in L) = (3.305*10-3 mole)/[(20.0 mL)*(1 L/1000 mL)] = 0.16525 mol/L = 0.16525 M
0.16525 M
0.16950 M
0.16960 M
Average molarity of acetic acid = 1/3*(0.16525 + 0.16950 + 0.16960) M = 0.16812 M ? 0.168 M (ans).
Grams of acetic acid per L (g/L):
Molar mass of acetic acid, HOAc = HC2H3O2 = (2*12.011 + 4*1.008 + 2*15.999) g/mol = 60.052 g/mol.
Grams of acetic acid per L = (0.168 mol/L)*(60.052 g/mol) = 10.0887 g/L ? 10.1 g/L (ans).
Mass percent of acetic acid in vinegar:
1 L vinegar solution contains 10.1 g acetic acid.
The density of the vinegar solution is 1.005 g/mL.
Therefore, mass of 1 L of vinegar solution = (1 L)*(1000 mL)/(1 L)*(1.005 g/mL) = 1005 g.
Thus, 1005 g solution contains 10.1 g acetic acid. Therefore, mass of solvent in the solution = (mass of solution) – (mass of acetic acid) = (1005 g) – (10.1 g) = 994.9 g.
Mass percent of acetic acid in vinegar = (mass of acetic acid)/(mass of solvent)*100 = (10.1 g)/(994.9 g)*100 = 1.015% (ans).
Sample 1
Sample 2
Sample 3
Moles of KHP (Molar mass = 204.2)
Moles = (grams of KHP)/(gram molar mass of KHP)
= (0.56 g)/(204.2 g/mol) = 2.7424*10-3 mole
2.7424*10-3 mole
2.6934*10-3 mole
2.6689*10-3 mole
Moles of the base required
The balanced chemical equation for the reaction is
NaOH (aq) + KHP (aq) --------> NaKP (aq) + H2O (l)
As per the stoichiometric equation,
Moles KHP = moles NaOH.
2.7424*10-3 mole
2.6934*10-3 mole
2.6689*10-3 mole
Molarity of NaOH
Molarity of NaOH = (moles of NaOH)/(volume of the base used in L)
= (27424*10-3 mole)/[(28.02 mL)*(1 L/1000 mL)] = 0.09787 mole/L ? 0.09789 M
0.09789 M
0.09972 M
0.10148 M