Here are the acid or base dissociation constants (whichever is applicable) for e

Question

Here are the acid or base dissociation constants (whichever is applicable) for each of the eight species: Calculate the pH of a 0.1 M solution of each of these substances. Simply write the results of your calculations in the table below; but attach all calculations when you turn in this prelab. Also transfer these results to the table in the lab report on page 9. In addition, predict using a calculation whether you would expect the bromocresol green indicator will appear yellow, green, or blue when added to the solution. (Recall that the indicator has K_1 = 1.3 times 10^-5.)

Explanation / Answer

(1): The balanced equation for the dissociation of the weak acid CH2ClCOOH is

---------------- CH2ClCOOH --------- > CH2ClCOO-(aq) + H+(aq) : Ka = 1.4x10-3

Init.Conc(M): 0.1 ---------------------- 0 ---------------------- 0

change(M) : - y ------------------------ + y ------------------- + y

Eqm.conc(M): (0.1 - y), -------------- y ---------------------- y

Ka = 1.4x10-3 = [CH2ClCOO-(aq)]x[H+(aq)] / [CH2ClCOOH] = y x y / (0.1 - y) =  y2 / (0.1 - y)

=>  y2 + 1.4x10-3 y - 1.4x10-4 = 0

=> y = 0.01115285 M

Hence [H+(aq)] = y = 0.01115285 M

=> pH = - log[H+(aq)] = - log(0.01115285 M) = 1.953 (answer)

Indicator color - yellow (answer)

(2): Similarly for the dissociation of CH3COOH

Ka = 1.8x10-5 = [CH3COO-(aq)]x[H+(aq)] / [CH3COOH] = y x y / (0.1 - y) =  y2 / (0.1 - y)

=>  y2 + 1.8x10-5 y - 1.8x10-6 = 0

=> y = 0.0013327 M

Hence [H+(aq)] = y = 0.0013327 M

=> pH = - log[H+(aq)] = - log(0.0013327 M) = 2.875 (answer)

Indicator color - yellow (answer)

(3):

Similarly for the dissociation of C8H9NH2

Kb = 5.6x10-4 = [C8H9NH3+(aq)]x[OH-(aq)] / [C8H9NH2] = y x y / (0.1 - y) =  y2 / (0.1 - y)

=>  y2 + 5.6x10-4 y - 5.6x10-5 = 0

=> y = 0.007209 M

Hence [OH-(aq)] = y = 0.007209 M

=> pOH = - log[OH-(aq)] = - log(0.007209 M) = 2.142

=> pH = 14 - pOH = 14 - 2.142 = 11.858 (answer)

Indicator color - blue (answer)

(4):

Similarly for the dissociation of Zn(H2O)42+(aq)

Ka = 1.0x10-6 = [Zn(OH)(H2O)3+(aq)]x[H+(aq)] / [Zn(H2O)42+(aq)] = y x y / (0.1 - y) =  y2 / (0.1 - y)

=> y = 3.1573x10-4 M

Hence [H+(aq)] = y = 3.1573x10-4 M

=> pH = - log[H+(aq)] = - log(3.1573x10-4 M) = 3.501 (answer)

Indicator color - yellow (answer)

(5):

Similarly for the dissociation of Al(H2O)63+(aq)

[H+(aq)] = y = 9.950x10-4 M

=> pH = - log[H+(aq)] = - log(9.950x10-4 M) = 3.022 (answer)

Indicator color - yellow (answer)

(6): Similarly for the dissociation of Na2CO3

[OH-(aq)] = y = 4.1536x10-3 M

=> pH = - log[OH-(aq)] = - log(4.1536x10-3 M) = 2.382 (answer)

=> pH = 14 - 2.382 = 11.618 (answer)

Indicator color - blue (answer)

(7) :

Similarly for the dissociation of NaHSO4

[H+(aq)] = y = 0.02916 M

=> pH = - log[H+(aq)] = - log(0.02916 M) = 1.535 (answer)

Indicator color - yellow (answer)

(8) Since HCl is completely dissociated,

[H+] = [HCl] = 0.1

=> pH = - log(0.1) = 1.00 (answer)

Indicator color - yellow (answer)

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