Chemistry Questions?

Question

a. Babituric acid, HC4H3N2O3, is used to prepare barbiturates, a class of drugs used as sedatives. A 325 mL aqueous solution of barbituric acid has a pH of 2.34 and contains 9 g of the acid. What is the Ka for barbituric acid?

b.Penicillin (MM = 356 g/mol), an antibiotic often used to treat bacterial infections, is a weak acid. Its Ka is 1.7 x 10^-3. Calculate [H+] in solutions prepared by adding enough water to following to make 1.30 L.

Answer 1

Lancenigo di Villorba (TV), Italy

Your question interests two WEAK ACID's SOLUTIONs.

ABOUT STOICHIOMETRY
W. Ostwald was the Study's Master of Arrhenius, so when the second chemist theorized the ions in order to correlate his experimental findings then the former developed the SIMPLEST MATHEMATICAL RELATION ABOUT WEAK ELECTROLYTE's SOLUTION.
Speaking about MONO-VALENT or MONO-BASIC WEAK ACID, Arrhenius was interested to the following Chemical Equilibrium

HA(aq) <---> H+(aq) + A-(aq)
Ka = |H+| * |A-| / |HA|

which it happens in aqueous media ; nonetheless, water itself undergoes to an Ionic Dissociation's Equilibrium

H2O(aq) <---> H+(aq) + OH-(aq)
Kw = |H+| * |OH-|

Ostwald did the assumptions in order to NON-DILUTED AQUEOUS SOLUTIONs of a MONOBASIC WEAK ACID, the following one :

|H+| = (Ka * |HA|)^0.5

since he retained that |HA| or the ACID's MOLARITY could remain unchanged.
Usually, we can use this formula in the case of MEDIUM-HIGH ACIDIC's STRENGHT.

##########################################
########## BARBITURIC ACID ################
##########################################
pH = 2.34
|H+| = ALOG(0 - pH) = 4.6E-3 M
|HA| = (Acid's Mass) / (Acid's Volume * MW) =
= 9 / (0.325 * 128) = 0.216 M
Ka = |H+|^2 / |HA| = (4.6E-3)^2 / 0.216 = 9.8E-5

##########################################
########## PENICILLIN ######################
##########################################
You gave not the Acid's Mass.
I know that usual PENICILLANIC ACID's Molarities stay around like 50 g/L, e.g. 0.14 M.
|H+| = (Ka * |HA)^0.5 = (1.7E-3 * 0.14)^0.5 = 1.5E-2 M
pH = 0 - LOG(|H+|) = 1.8

I hope this helps you.

Answer 2

Let's set it up. You can grind through the numbers. (1) Find the mol wt of the acid.
Divide 9g by that weight to get the moles.
Multiply the moles by 1000/325 to get the molar conc in moles/L. Now convert the pH to an [H+] conc. [H+]= 10^-2.34 = 10^-3 + 10^.67 which is about 4.8x10-3.
Now the dissocation of the acid in water reaction is acid -> H+ + (barbituate ion)-
If the acid is the only source of each, the conce of the barbituate ion will also be about 4.8x10-3.
Then Ka= (4.8x10-3^2)/(moles acid added-4.8x10-3)

Your penicillin problem is missing some information.