Question

In: Chemistry

Weight of Unknown #4( in g) Volume of water (in mL) Unknown #4 solution (in g/L)...

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.

Solutions

Expert Solution

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.


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