In: Chemistry
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.
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.