Question 1:For many weak acid or weak base calculations, you can use a simplifyi

Question

Question 1:For many weak acid or weak base calculations, you can use a simplifying assumption to avoid solving quadratic equations.Classify these situations by whether the assumption is valid or the quadratic formula is required.

HA=.001 Ka=1x10-5

HA=1M Ka = 1x10-3

HA=.01M Ka=1x10-5

HA=.1M Ka=1x10-3

HA=.01M Ka = 1x10-4

Question 2: The Ka of a monoprotic weak acid is 5.20 × 10-3. What is the percent ionization of a 0.189 M solution of this acid?

Question 3: Phosphoric acid, H3PO4(aq), is a triprotic acid, meaning that one molecule of the acid has three acidic protons. Estimate the concentrations of all species in a 0.350 M phosphoric acid solution. pKa1=2.16 pKa2=7.21 pKa3=12.32

H3PO4 =

H2PO4- =

(PO4)3- =

H+ =

Explanation / Answer

Question 1 )

only two are valid

HA=1M, Ka = 1x10-3

[H+] = x = sqrt (Ka x C) = sqrt ( 10^-3 x 1 ) = 0.0316

percent dissociation = x x 100 / C = 0.0316 x 100 / 1 = 3.16%

it is < 5%   . so it is valid

HA=.01M, Ka=1x10-5

x = sqrt (10^-5 x 0.01) = 3.16 x 10^-4

percent dissociation = x / C ) x 100 = 3.16 x 10^-4 x 100 / 0.01 = 3.16 %

it is < 5%   . so it is valid

question 2 )

HA ---------------------> H+ + A-

0.189-x                        x        x   ---------------------> equilibrium

Ka = [H+][A-]/[HA]

Ka = x^2 / 0.189 -x

5.20 x 10^-3 = x^2 / 0.189 -x

x^2 + 5.20 x 10^-3 x - 9.83 x 10^-4 = 0

x = 0.0289

percent ionisation = x x 100 / C

                      = 0.0289 / 0.189 ) x 100

                      = 15.29 %

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