Question

U Crone (NH4CI). What is the pH of the solution? 3. (4 pts) Calculate the pH of a 0.250 M solution of potassium phenolate, KC6H50. Ka foi phenol (C6H5OH) is 1.0 x 10-10. 4. (8 pts) How do the concentration/volumes of the buffer affect the buffer capacity! E.g. 50.0 mL of 0.10 M acetic acid solution with 50.0 mL of 0.10 M sodium acetate solution vs. the buffer you made in the lab (25.0 mL of 0.10 M acetic acid solution with 25.0 mL of 0.10 M sodium acetate solution).

Answer 1

solutions we know that potassium phenalate KC6H50 react with water H₂O to porte soention. XC6 HO + H₂O - C6HBOH + KOH - KC6H5O + H2O Initial 0.250M Final o C6H5OH + koh 0 0 0.250M 0.250M Given concentration of [xC6Hgo] = 0,250 and ICC6H5O is completely dissociate into water. so, concentration of CongoH = 0.250M (C6H5OH) = 0.250m For C6H5OH, dissociate not completely CoHoo + H+ C6H5OH Initial 0.25om Change an equilibriurf 0.250-2) M tr th 2 we know that - Ica = [C6H30] [H + (С Нѕон)

ka = (2) (c) 0.25-2 Given ka= 1.0x10 10 1.0x100 = 2 0.25-2 1.0 x10 to this value is very small so n will be also very as compare to 0.25. So, neglect from denominator. So we can write -10 10x10 = 25 n=√2.5 810-11 n= 1 - = 5x10-6 m 200000 n=58106m concentration of Ht at equilibrium = a so, [H+] = 5x106m

We know that - (pH = -logso [ke pH = -logoo (5x1056 m) pH = -logso (5) - Bogoo (109) pH = 5.3 . -30.01 ᎯᎠ .
this is the solution of question no. 3 only.