Question

< Question 5 of 10 > Attempt 7 - Determine the pH of a solution containing 0.050 M NaOH and 0.045 M KI neglecting activities. pH = 12.7 Determine the pH of the same solution including activities. Activity coefficients at various ionic strengths can be found in this table. pH= about us careers privacy policy terms of use contact us help

I need the pH that includes the activities.

Answer 1

We can solve given problem in following steps

Step 1 : Calculation of 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 , [ NaOH ] = [ Na ⁺ ] = [ OH ^- ] = 0.050 M , [ KI ] = [ K ⁺ ] =  [ I ^- ] = 0.045 M

Ionic strength () of solution = 1/2 [ ( 0.050 ) ( +1 ) ² + ( 0.050 ) ( -1) ² + ( 0.045 ) ( +1) ² + ( 0.045) ( -1) ²  ]

= 1/2[ 0.050 + 0.050 + 0.045 + 0.045 ]

= 1/2 [ 0.190 ]

= 0.0950 M

Step 2 : Calculation of activity coefficient of OH ^-

We have Debye Huckel equation : log = ( - 0.51 z ²) / ( 1 + ( / 305)

Where is activity coefficient, z is charge on ion, is ion size and is ionic strength.

For OH ^- , z = - 1 , = 350 pm and = 0.095 M

log = ( - 0.51 (- 1) ²0.095 ) / ( 1 + ( 350 0.095 / 305 ) )

log = ( - 0.51 (1)( 0.3082 ) / ( 1 + ( 350 (0.3082) / 305 ))

log = - 0.1572 / ( 1 + 0.3537 )

log = - 0.1572 / 1.3537

log = - 0.116

= 10 ^{- 0.116}

= 0.766

Step 3 : calculation of pOH of solution.

We have relation, pOH = - log [ OH ^- ] _OH ^-

pOH = - log ( 0.050 $\times$ 0.766 )

pOH = - log 0.0383

pOH = 1.455

Step 4 : Calculation of pH of solution.

We have relation, pH + pOH = 14

pH = 14 - pOH

pH = 14 - 1.455

pH = 12.545

ANSWER : pH of solution = 12.5