Question

Calculate the pH of a solution containing 0.016 M NaOH plus 0.0120 M LiNO3 using activities and Neglecting activities. this is for a quantitative analysis class and i keep getting 11.3 for both i must be doing it wrong.. pleas sig fig the answers for me

Answer 1

Ionic Strength = 1/2*(0.016*1 + 0.012 * 1 + 0.019 *1 + 0.012) = 0.0295

IS = 0.0295 M

sso..

including activities

Kw = γH⁺[H⁺]·γOH^-[OH^-]

solve for  γH⁺[H⁺]

γH⁺[H⁺] = Kw / γOH^-[OH^-] = (10^-14)= / (γOH*0.016)

we need to appl y debye-Huckel equation

activity of OH-:

log(γOH) =  [-0.51·(-Zi)²·(IS)^0.5]/[1+350·(Zi)^0.5/305]

alpha = 350 pm

log(γOH) = (-0.51*1*(sqrt(0.0295 )) / ( 1+ 900 * sqrt(0.0295)/305))

log(γOH) = -0.0581326

γOH = 10^-0.0581326 = 0.87471

so..

γH⁺[H⁺] = Kw / γOH^-[OH^-] = (10^-14)= / (0.87471*0.016) = 7.145*10^-13

then

pH = -log( γH⁺[H⁺]) = -log( 7.145*10^-13) = 12.1459

pH = 12.1459