Weight of Unknown #4( in g) Volume of water (in mL) Unknown #4 solution (in g/L)
ID: 877746 • Letter: W
Question
Weight of Unknown #4( in g)
Volume of water (in mL)
Unknown #4 solution (in g/L)
0.118
50.00
2.36
0.127
50.00
2.54
0.112
50.00
2.24
Volume of NaOH used = Final concentration – Initial concentration
= 16.49 mL – 4.90 mL = 11.49mL NaOH= 0.01149 L NaOH
Exp. #
Concentration of NaOH
Initial volume of NaOH
Final volume of NaOH
Volume of NaOH used for titration
1
0.1030M
4.90mL
16.49mL
11.59mL
2
0.1030M
16.49mL
26.01mL
9.52mL
3
0.1030M
26.01mL
35.09mL
9.08mL
a) Propose whether the unknown #4 is monoprotic or diprotic.
b) Propose the molar mass of the unknown #4.
c) show the calculation for the assumption for the monoprotic or diprotic.
Weight of Unknown #4( in g)
Volume of water (in mL)
Unknown #4 solution (in g/L)
0.118
50.00
2.36
0.127
50.00
2.54
0.112
50.00
2.24
Explanation / Answer
For acid (unknown #4):
weight = 0.118g
volume = 50ml
For Base (NaOH):
weight = ?
molar mass of NaOH:40
volume used=11.49ml
concentration(M) = 0.103M
For titration: Macid X Vacid = Mbase X Vbase
Therefore Macid = (Mbase X Vbase)/ Vacid= (0.103x11.49)/50 = 0.023M
concentration of acid (unknown #4) = 0.023M;
Molarity = (weight/Molar mass) x (1000/volume in ml)
Molar mass = (weight/Molarity) x (1000/volume in ml)
= (0.118/0.023)x(1000/50)
= 102.6
Molar mass of acid = 102.6 ;
number of moles of acid (n) = weight / molar mass = 0.118/102.6 = 0.00115 moles
Number of moles of NaOH = weight/Molar mass
To find out weight of NaOH,
Molarity = (weight/Molar mass) x (1000/volume in ml)
weight = (Molarity x molar mass x volume)/1000
= (0.103 x 40 x 11.59)/1000
= 0.047grams
Then number of moles = weight/Molar mass = 0.047/40 = 0.00119
number of moles of base/number of moles of acid = 0.00119/0.00115 = 1.03 that is almost equal to 1
That means one mole of base is reacting with exactly one mole of acid. It is possible only if the acid is monoprotic.