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.

Answer 1

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:   M_acid X V_acid = M_base X V_base

Therefore M_acid = (M_base X V_base)/ V_acid= (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.