What is the molar mass of the acid?

Question

A sample of 0.1341 g of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.0950 M NaOH. The acid required 19.3 mL of base to reach the equivalence point.

(a)What is the molar mass of the acid?

(b) After 12.0 mL of base had been added in the titration, the pH was found to be 5.10. What is the Ka for the unknown acid?

Answer 1

(a)
You have stated that the unknown is a monoprotic acid. The neutralisation equation is

HA + NaOH ===> NaA + H2O

Assume HA has a MW of M. From stoichiometry

0.1341 / M = 19.3 x 0.095 / 1000

Solving for M leads to M = 73.1

Molar mass 73.1 g

(b)
The titration curve is governed by the following charge balance

[Na+] + [H+] = [OH-] + [A-]

Now [Na+] = 12 / (12 + 25)*0.095 = 0.0308 mol/l

Given that [H+] will be << [Na+] and similarly [A-]will be >> [OH-], it can be safely assumed that [Na+] = [A-].

Just so there is no unjust assumptions check
[H+] = 10^-5.1 = 7.94x10^-6 mol/l

From Kw (1x10^-14) [OH-] = Kw/[H+] = 1.26x10^-9 mol/l

OK, continuing

[HA] + [A-] = 0.1341 x 1000 /(25 +12 ) / 73.1 = 0.0496 mol/l

Therefore [HA] = 0.0496 - 0.308 = 0.0188 mol/l

Rearranging Ka leads to

pKa = pH - log([A-]/[HA]) = 5.1 - log(0.308/0.188) = 4.88

or Ka = 1.30x10^-5

Answer 2

Lancenigo di Villorba (TV), Italy

Well, stay tuned while I execute your numerical requests.

Question a)

The alkalimetric equivalents which you used are :

0.095 * 19.300 = 1.8335 mmol of NaOH

they are also like 1.8335 meq of NaOH, hence 1.8335 meq of acid. You said me the acid are monoprotic, hence you manipulated 1.8335 mmol fo ACID.
Thus, an acid's mass like 0.1341 g constitued itself by 0.0018 acid's mol. You calculate molecular weight (or molecular mass) of acid, as follows :
0.1341 / 0.0018 = 73 (g / mol)

Question b)

You said me that, during alkalies titration, at one point you can measure pH's titration's mixture as 5.1.
At this point, hydrogen concentration is rescued by pH's value
ALOG(0-5.10) = 0.0000794 (mol / lt)
concentration which derived by ions balancing of a well-known acido-base reaction (H+ + OH- <---> H2O).
I estimate the original amount of hydrogen ions by adding mmol of alkalies used at measure's point to mmol of hydrogen ions present at the same point, as follows
0.00000794 * (25.000 + 12.000) + 0.0950 * 12.000 = 1.140 (measure's unit as mmol)
At the original point, hydrogen ion's concentration
1.140 / 25.000 = 0.046 (mol /lt)
At the original point, acid's concentration (total of undissociate and dissociate forms)
1.8335 / 25.000 = 0.073 (mol /lt)
At the original point, dissociate form's concentration by molar balancing of acid
0.073 - 0.046 = 0.027 (mol /lt)
By adapting the well-known Ostwald's formula
(0.046) * (0.046) / (0.027) = 0.075 (mol /lt)
So, this is the constant's acidity which you search.
pKa' s value is 1.12.

I hope this helps you.

Answer 3

HA + NaOH ---> H2O + NaA

the moles of NaOH : 0.095 * 0.00193 = 1.83*10^-3
the moles of HA = the moles of NaOH
the molar mass of the acid 0.1341 / 1.83*10^-3 = 73.14

the moles of 12ml of base 0.095 * 0.0012 = 1.14*10^-3
the moles of HA left after reacting with 12ml of base : 1.83*10^-3 - 1.14*10^-3 = 6.9*10^-4

HA <---> H+ + A-

the moles of H+ produced : Ka*moles of HA = Ka*6.9*10^-4
the molar concetration of H+ in the solution = [H+] / V = Ka * 0.0187

pH = -log[H+] = -log(Ka*0.0187) = 5.1
=> Ka *0.0187 = 10^(-5.1) => Ka = 4.25*10^-4