15 mL of 5% NaOH (w/v) is used to deprotonate the unknown carboxylic acid. The T

Question

15 mL of 5% NaOH (w/v) is used to deprotonate the unknown carboxylic acid. The TBME is washed with an additional 10mL of the NaOH solution and then 20 mL of water which are all combined. Later, 10 mL of 6.0M HCl was used to protonate the carboxylate ion. Assume the concentration of carboxylate is negligible.
                                 

                    a) After the carboxylate is protonated, how many moles of HCl are left in excess?                 

                    b) What is the solution's final H+ ion concentration?                 

                                     c) What is the pH of the final solution?

Explanation / Answer

5 % w/v NaOH means 5 gm per 100 ml solution

for deproetonation amount of NaOh used = ( 15 x 5/100) = 0.75 gm

NaOH moles used = mass/molar mass = 0.75/40 = 0.01875,

at neutralisation NaOH moles = moles of H+ = moles of acid , hence moles of acid = 0.01875

now additional 10 ml NaOH is added hence additional OH- ions = (10 x5/100) = 0.5 gm

NaOH moles = 5/40 = 0.0125,

moles of H+ added = MV = 6 x (10/1000) = 0.06

a) moles of HCl required for protonation of carboxylation = moles of NaOH used

= 0.01875+0.0125 = 0.03125

excess HCl left = ( 0.06-0.03125) = 0.02875

b) solutional final H+ =0.02875 moles

volume = 15+10+20+10 = 55 ml = 0.055 liters

[H+] = ( 0.02875/0.055) = 0.523

c) pH = -log [H+] = -log ( 0.523) = 0.2815

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