Determine the pH of a solution containing 0.040 M NaOH and 0.025 M KI... A table

Question

Determine the pH of a solution containing 0.040 M NaOH and 0.025 M KI... A table of activity coefficients at various ionic strengths can be found here ...neglecting activities Number 12.60 including activities Incorrect. You have correctly found the pH of the solution neglecting the activities. To find the pH including activities, first find the ionic strength of the solution Number 11.92 where c is the concentration of the ith species and z is its charge. Look in your text to find the activity coefficient for the OH ion at the calculated ionic strength. You will have to use linear interpolation to find YoH Next, find the activity of H* in the solution. Then use the activity of H to find the pH. pHlog H

Explanation / Answer

First, get IS

Recall that ionic strength considers all ions in solution, and its charges. It is typically used to calculate the ionic activity of other ions. The stronger the electrolytes, the more ionic strength they will have.

The formula:

I.S. = 1/2*sum( Ci * Zi^2)

Where

I.S. = ionic strength, M (also miu / ) used

Ci = concentration of ion “i”

Zi = Charge of ion “i”

The exercise:

IS =1/2 * ((0.04) + 0.04 + 0.025 + 0.025) = 0.065

now,

get

activitiy coefficeints:

Recall that:

-log() = 0.51*(Zi^2)*sqrt(I.S.) / ( 1 + ( * sqrt(I.S)/305))

Where

i = activity coefficient for species “i”

i = theoretical diameter in pm (10^-12 m)

Zi = Charge of ion

I.S. = ionic Strength (usually used as as well)

If we wanted only

= 10^-(0.51*(Zi^2)*sqrt(I.S.) / ( 1 + ( * sqrt(I.S)/305)))

for OH- , Zi = -1 IS = 0.065, = 350

YOH- = 10^-(0.51*(1^2)*sqrt(0.065) / ( 1 + (350* sqrt(0.065)/305))) = 0.793

for H+- , Zi = -1 IS = 0.065, = 900

YH+- = 10^-(0.51*(1^2)*sqrt(0.065) / ( 1 + (900* sqrt(0.065)/305))) = 0.8429  

then

Activity of H+ = Kw/([OH-]Y-OH) = (10^-14)/(0.04*0.793 ) = 3.152*10^-13

pH = -log(Activity of H+ ) = -log( 3.152*10^-13) = 12.501

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.