Question

i keep getting 1 wrong.. its says the total V is hcl and water added together but i did it and its not right. thanks

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 ___________________

Answer 1

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

Solution :- we have volume and molarity of the HCl solution and also know the volume of the water used for the dilution

Therefore lets first calculate the new molarity of the HCl solution at total volume

Total volume = 27.81 ml + 1.44 ml = 29.25 ml

New molarity of HCl= initial moalrity of HCl * initial volume / total volume

=1.848 M * 1.44 ml / 29.25 ml

= 0.090978 M

Now lets calculate the pH using this concentration

pH= - log [H3+ ]

pH = - log [0.090978]

pH = 1.041

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

Solution :- acid is monoprotic therefore it has 1 :1 mole ratio with the NaOH

Lets first calculate the moles of the NaOH using its volume and molarity

Moles of NaOH = 0.2213 mol per L * 0.01380 L =0.003054 mol NaOH

Therefore moles of acid reacted = 0.003054 mol

We only used half of the original solution for the titration therefore total moles of the acid present = 0.003054 mol * 2= 0.006108 mol

a)Using this moles of acid lets calculate the molar mass of the acid

Molar mass of acid = mass of acid / moles of acid

= 1.8105 g / 0.006108 mol

= 296.4 g per mol

b)Now molarity of the unknown acid from titration = moles of acid / volume of acid

= 0.003054 mol / 0.025 L

= 0.1222 M

c)Ka of unknown acid

When the solution from the titration was added to original half solution then pH is 3.24

After adding the titrated solution the moles of acid and its conjugate base are same therefore pH = pka at equivalence point.

Therefore 3.24 is the pka of the acid

Now lets convert pka to ka of acid

Pka= -log ka

Therefore

Ka = antilog [- pka]

Ka= antilog[ -3.24]

Ka= 0.000575

Now lets calculate the undissociated acid from pH measurement

pH of orginal acid solution is 2.13

therefore lets calculate the H+ concentration using this pH value

pH= -log[H+]

therefore [H+] = antilog [- pH]

= antilog[- 2.13]

= 0.007413 M

d)Now lets calculate the concentration of the undissociated acid

Concentration of the undissociated acid = original concentration – [H+]

= 0.1222 M – 0.007413 M

= 0.1148 M

e) concentration of unknown acid from pH

pH= -log[H+]

therefore [H+] = antilog [- pH]

= antilog[- 2.13]

= 0.007413 M