Question

Calculate the pH of a solution which is 0.0070 M in KOH and 0.019 M in NaNO3 using activities. Submit Answer Tries 0/5 Calculate a pH for the same solution, ignoring activities. Submit Answer Tries 0/5

Answer 1

Given

concentration of KOH = 0.0070 M

concentration of NaN03 = 0.019 M

the ions percent in KOH is K­­⁺ and OH^-

the ions percent in NaNO3 is Na­­⁺ and NO_­3­^-

the calculations of activities is calculated using the formula

μ = 1/2 (c1 Z1² + c2 Z2² +..........)

= 1/2(0.0070 (+1)² + 0.0070(-1)² + 0.019(+1)² + 0.019(-1)²)

= 0.026 M

γ_­(OH)­ activity coefficient =0.873

K­_W­ =A­_H­ A_OH­

A­_H­ = K­_W­ / (A_­OH X ­ γ_­(OH­)

= 10^-14 / (0.0070 X 0.873) = 1.63 X 10^-12

pH = - log[A_­H­] = -log(1.63 X 10^-12)

pH = 11.787

(B)

pH after neglecting activities

[OH^-] =0.0070

K_­W­ = [H_­⁺]­ [OH^-]

[H⁺] = K_­W­ /[OH^-]

[H⁺] =10^-14 / 0.0070

          = 1.428 X 10^-12

pH = - log[H⁺]

= - log(1.428 X 10^-12)

pH = 11.845

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