Question

Question 9 of 11 > 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 solution including activities. Activity coefficients at various ionic strengths can be found in this table. pH =

Answer 1

pH of solution neglecting activities

We have, [NaOH] = 0.050 M

NaOH dissociates completely into water, hence [NaOH ] = [OH ^- ] = 0.050 M

We have relation, [H⁺] [OH ^-] = 1.0 $\times$ 10 ^-14

$\therefore$ [H⁺] = 1.0 $\times$ 10 ^-14 / [OH ^-] = 1.0 $\times$​​​​​​​ 10 ^-14 / 0.050 = 2.0 $\times$ 10 ^-13 M

pH = - log [H⁺]

$\therefore$ pH = - log 2.0 $\times$ 10 ^-13 = 12.6989 = 12.70

ANSWER : pH of a solution containing 0.050 M NaOH and 0.050 M KI is 12.70

pH of solution considering activities

First we need to calculate ionic strength of solution.

We have, Ionic strength () = 1/2 Sum ( C _i Z _i² )

Where C _i is concentration of i th species and Z _i is its charge.

Ionic strength of solution = 1/2 [ [Na ⁺] (+1) ² + [OH ^- ] (-1) ² + [K ⁺] (+1) ² + [I ^-] (-1) ² ]

We have, [Na⁺] = 0.050 M , [OH ^- ] = 0.050 M and [K ⁺] = 0.050 M, [I ^-] = 0.050 M

Hence, Ionic strength of solution = 1/2[ 0.050 (+1) ² + 0.050(-1) ² + 0.050 (+1) ² + 0.050 (-1) ²]

= 1/2 [ 0.050 + 0.050 + 0.050 + 0.050 ]

= 1/2 [ 0.20 ]

= 1/2 [ 0.20]

= 0.10 M

Referring to literature, for = 0.10 M   OH ^- = 0.76

We have relation, ([H⁺] H ⁺ ) ([OH ^- ] OH ^-) = K _w = 1.0 10 ^-14

Therefore, ([H⁺] H ⁺ ) = 1 10 ^-14 / ([OH ^- ] OH ^-) = 1.0 10 ^-14 / 0.050 ( 0.76) = 2.63 10 ^-13

We have, pH = - log ([H⁺] H ⁺ ) = - log 2.63 10 ^-13 = 12.578

ANSWER : pH of a solution containing 0.050 M NaOH and 0.050 M KI = 12.58