1. You add Hydrochloric Acid to distilled water. If all the HCl dissociates in s

Question

1. You add Hydrochloric Acid to distilled water. If all the HCl dissociates in solution, calculate the pH of the final solution given the following data. (assume the liquid volumes are addative)
Volume of distilled water used 27.81 mL
Volume of HCl added 1.44 mL
Concentration of added HCl 1.848 M

pH of diluted HCl solution ___________________

2. You weigh a sample of a monoprotic unknown acid and dissolve it in 50.00 mL of distilled water. Exactly half of this solution is titrated with Sodium Hydroxide to the phenolphthalein end point.
The pH of the other half of the original solution is measured with a pH meter. The "neutralized" solution is added to the "original" solution and the pH of this combined "final" solution is also measured.
The following are the measured values:
Mass of unknown acid 1.8105 g
Volume of NaOH used in titration 13.80 mL
Concentration of the NaOH used 0.2213 M
pH of the original acid solution 2.13
pH of the final acid solution 3.24

      CALCULATE the following
(a) Molecular Weight of Acid used in titration ___________________

(b) Molarity of UNKNOWN Acid solution from titration ___________________

(c) Ka of UNKNOWN Acid ___________________

(d) Concentration of undissociated Acid from pH measurements ___________________

(e) Total concentration of UNKNOWN acid from pH measurements ___________________

Explanation / Answer

1. No of moles of HCl =1.848*(29.25/1000)

   = 0.054 mole
pH = -log [H3O+] = -log 0.054 = 1.268

2. pH of the original acid solution = 2.13

pH = -log[H3O+]

[H3O+] = 10^(-pH) = 10^(-2.13) = 0.0074 M

as it is monoprotic acid , No of moles of H3O+ = ACID

Concentration of undissociated Acid from pH measurements = 0.0074

Total concentration of UNKNOWN acid from pH measurements   = 0.0148(1000/50) = 0.296 M

Ka of UNKNOWN Acid =

from dilution formula M1V1 = M2V2

M1V1 = no of moles of acid = 0.003 mole

M2V2 = NO of moles of base= (0.2213*13.8) /1000 =

[H3O+] = C*X   = 0.003

c = 0.0074 x = degree of dissociation = ?

x = 0.003/0.0074 = 0.4

Ka = c*x^2 = 0.296*(0.4)^2 = 0.04736

Molecularweight of acid = (1.8105/0.296)*(1000/50)

   = 122.33 g/mol

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