Question

Determine the pH of a solution containing 0.050 M NaOH and 0.025 M KI neglecting activities. pH= Determine the pH of the same
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Answer #1

(1) Neglacting Activities :- [OH-] = 0.050 M

POH = - log[OH-] = 1.30

PH + POH = 14

PH = 14 - 1.30

PH = 12.7

Ionic strength (μ) = 1/2 *(μ1Z12 +  μ2 Z22 +   μ3 Z32 +   μ4 Z42.)

= 1/2 * (0.035 X 12 + 0.035 X 12 + 0.025 X 12 + 0.025 X 12)

= 0.06 M

μ = 0.05 rOH- = 0.81

μ = 0.1 rOH- = 0.76

μ = 0.06 rOH-​​​​​​​ = ?

Linear inter portion rOH- = 0.81 + (0.76 - 0.81) . (0.06 - 0.05) / (0.1 - 0.05)

formula y = yo + [(y1 - yo) (x-xo) / (x1 - xo)]

rOH- = 0.8 M

[H+]rH+ = KW / ( [OH-] rOH-) = 10-14 / (0.035 X 0.8) = 3.57 X 10-13 M

pH = - log [H+] = = log(3.57 X 10-13 M)

pH = 12.45

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Answer #2
  1. 12.70

  2. 12.63


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