so confused?

Question

cyanic acid is a weak monoprotic acid if the intial concentration of cyanic acid is 0.150 M and the equilibrium concentration of H^3O+ is 4.8 x 10^-2 M, calculate ka for cyanic acid

isobutylamine is a weak base if the initall concentration of isobutylamine is 0.055 M and the equilibrium concentration of OH- is 4.0 x10^-3 M. calculate kb for isobutylamine

Gallic acid is a weak monoprotic acid If the initial concentration of gallic acid is 0.280 M and the equilibrium concentration of H^3O+ is 3.3 x 10^-3 M, calculate ka for galic acid

Trimethylamine is a weak base If the initial concentration of trimethylamine is 0.390 M and the equilibrium concentration of OH- is 4.4 x 10^-3 M. calculate kb for trimethylamine

Dripropylamine is a weak base if the intial concentration of dipropylamine is 0.160 M and the equilibrium concentration of OH- is 0.012 M, calculate Kb for dipropylamine

need the problem done or examples please!

Answer 1

A weak monoprotic acid (just call it HA)
HA concentration is 0.150M
HA <===> H+ and A- (note: H+ is the same conc. as A-)
Ka = [H+] x [H+] / [0.150- 4.8 x 10^-2] Note: 0.150 M to start, but .048 moles ionized.
Solving this-
[H+] = 4.8 x 10^-2M
Ka = (4.8 x 10^-2) x (4.8 x 10^-2) / 0.102
Ka = 2.304 x 10^-3 / 0.102 = 1.92 x 10^-2

You can try doing the rest of the problems. The only way you can learn to do them is to DO THEM. Watching someone else solve them just isn't the same.

Answer 2

So all of these problems work the same way, i'll illustrate how to do the first one and then i'll let you figure out the rest. So the first step of any equillibrium problem is writing a balance equation, for the dissocation of cyanic acid, this is as follows:

CHNOH (cyanic acid) ----> H + CHNO

So, the next step is to write the expression (equation) for the equilibrium constant, Ka. All equilibrium expressions are always [Products]/[Reactants] where pure liquids and solids are excluded!

Ka = [products]/[reactants] = ([H][CHNO])/[CHNOH]

They tell us that the inital concentration of cyanic acid is 0.150 M and that the equilibrium conc of H3O (same as H) is 4.8*10^-2 M, so lets set up a table, where I stands for inital conc., C stands for change, and E stands for equilibrium concentration.

CHNOH (cyanic acid) ----> H + CHNO
I 0.150 M 0 M 0 M
C -x +x +x
E 0.150 - x 4.8 *10^-2 4.8*10^-2

So we know that the equilibrium conc. of H is 4.8*10^-2 and since we don't start off with any, that means x must be 4.8*10^-2, therefore, CHNO must also have an equilibrium concentration of 4.8*10^-2 and cyanic acid has an equilibrium concentration of 0.150 - 4.8 * 10^-2 = 0.102 M

So now we've got the equilibrium concentration of all of the species, and so we can plug these numbers into our equilibrium constant (Ka) expression and calculate it out

Ka = ((4.8*10^-2)*(4.8*10^-2))/.102 = 0.0226

Thus, we've solved the problem. All of the other problems are done in the same way, just be careful when writing your equilibrium expression because stoichiometric coefficents do matter!!

Answer 3

These are variations on the same problem.

If you have a weak MONOPROTIC acid or base, Ka or Kb respectively is given by

x^2/M = Ka or Kb
where x is the conc of [H3O+] for the acid or [OH-]
for the base, and M is the initial conc of the acid or base.