Question

3 attempts left Check my work Enter your answer in the provided box. Determine the pH at 25°C of a solution prepared by dissolving 0.0650 mole of sodium acetate in 1.00 L of 0.250 M acetic acid. (Assume that the addition of sodium acetate does not change the volume of the solution.)

Answer 1

Kindly look at the step- wise solution given below:- I hope it will be helpful for you.

And a host write the equation for the lonizations of acetic aid & the Ka expression Сиз Соон 49) +н, ос) - на о ФС) + Сп, сода Ka = [Chacoo -] [1300 ] [cytool] [120 6 ] = ka [Chalon - [ch cool Determine the no. of moles of acetic and No of moles of CH₂ Cool Conc. x vol in übe = 0125M X 1L = 0.25 mol Moles of Chacoond = 010 650 mol. ka = 118 x 10-5 Substitute , we set these values in tam [H 3 ] = 118810-5x 0.25 0.0 65 [1300] = 6.92X 10-5 pH = -log [m3o8] = -log (6.92X - 5) pH = 4.159 TpH = 4,16