Question

ale an ICE table and substitute the experimental Ka value for the unknown in Equation 3 to determine the pH of a 0.0020 M solution of the unknown (requires use of the quadratic equation) Show calculation (4 pts.): Exp Ka=2.8x10-3 Molarity = 0,11814 _(2 pts.) 0.118 X 1-X 1+X You must turn in these two pages along with the two graphs generated using a graphing program such as Excel. (20 pts.)

Answer 1

Answer:

Let an unknown acid is HA. with [HA] = 0.1181 M and Ka = 2.8*10^-3 Ionizes as,

HA (aq) <---> H⁺ (aq) + A^- (aq)

Let at equilibrium "X" M of HA ionizes,

ICE table is,

HA (aq) <---> H⁺ (aq) + A^- (aq)

Initial 0.1181 0 0

Change -X +X +X

At Eqm (0.1181-X) X X

Ka is given as,

Ka = [H⁺][A^-] / [HA] = 2.8*10^-3.

Placing equilibrium concentrations,

(X)(X) / (0.1181-X) = 2.8*10^-3.

X² / (0.1181-X) = 2.8*10^-3.

X² = (0.1181-X) * 2.8*10^-3.

X² + 2.8*10^-3 X = 0.1181 * 2.8*10^-3.

X² + 2.8*10^-3 X = 3.31*10^-4.

Add, third term = [(1/2)*(2.8*10^-3)] = 0.0196*10^-4. to both sides,

X² + 2.8*10^-3 X + 0.0196*10^-4.= 3.31*10^-4 + 0.0196*10^-4.

X² + 2.8*10^-3 X + 0.0196*10^-4.= 3.33*10^-4 .

(X+1.4*10^-3)² = 3.33*10^-4 .

Taking square root of both sides,

X+1.4*10^-3 = 1.8*10^-2 . ....... (Only +ve root taken as -ve root will give -ve answer which is not acceptable)

X = 1.8*10^-2  -1.4*10^-3.

X = 0.017

By ICE table,

[H⁺] = X = 0.017 M

pH = -log[H⁺] = -log(0.017) = 1.77

pH of this solution is 1.77

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