1O points). Hydrogen cyanide is a weak monoprotic acid with = 6.2 x 10-10. If 25

Question

1O points). Hydrogen cyanide is a weak monoprotic acid with = 6.2 x 10-10. If 25.00 mL of 0.1039 M hydrogen cyanide is titrated with 0.05994 M NaOH, which of the following indicators would be best for determining the endpoint of the titration? Only answers supported by proper calculations will receive full credit the answer alone is worth 1 point. 5. ( K, Indicator Bromocresol purple5.2 Transition pH Acid color Base colo Yellow Purple Blue 6.0- 7.6-92 Yllow Purple Cresol purple Thymol phthalein 8.3 10.5 Alizarin yellow Tropaeolin O Yellow Colorless Blue 10.1-12.0 Yellow Orange-red |Yellow 1.1- 12.7 |Orange

Explanation / Answer

Calcualte pH in equilbirium

mmol of acid = MV = 0.1039*25 = 2.5975

mmol of base required = 2.5975

V base = mmol/V = 2.5975/0.05994 = 43.3350

V total = 25+43.3350 = 68.335 mL

[CN-] = mmol/V = 2.5975/68.335 = 0.038011 M

in equilbirium

expect hydrolysis

CN- + H2O <-> HCN + O

since A- is formed

the next equilibrium is formed, the conjugate acid and water

A- + H2O <-> HA + OH-

The equilibrium s best described by Kb, the base constant

Kb by definition since it is an base:

Kb = [HA ][OH-]/[A-]

Ka can be calculated as follows:

Kb = Kw/Ka = (10^-14)/(6.2*10^-10) = 1.6*10^-5

get ICE table:

Initially

[OH-] = 0

[HA] = 0

[A-] = M

the Change

[OH-] = + x

[HA] = + x

[A-] = - x

in Equilibrium

[OH-] = 0 + x

[HA] = 0 + x

[A-] = M - x

substitute in Kb expression

Kb = [HA ][OH-]/[A-]

1.6*10^-5 = x*x/(0.038-x)

solve for x

x^2 + Kb*x - M*Kb = 0

solve for x with quadratic equation

x = OH- =7.717*10^-4

[OH-]  =7.717*10^-4

pOH = -log(OH-) = -log(7.717*10^-4 = 3.11

pH = 14-3.11= 10.89

best will be

Alizarin yello, which changes from 10.1 to 12

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