The pH is determined by the hydrolysis of the salt.
M of the salt is 0.05/0.150 = 0.3333 (you can round to the right number of s.f. as the last step.)
............OCN^- + HOH ==> HOCN + OH^-
initial...0.3333.............0......0
change.....-x................x.......x
equil....0.3333-x............x.......x
Kb for OCN^- = (Kw/Ka for HOCN) = (HOCN)(OH^-)/(HOCN)
Substitute and solve for x = OH^- and convert to H.
calculate the ph of a solution that results from titrating 0.0500 moles of cyanic acid to the equivalence point with an equal number of moles of OH-, if the final volume of the solution is 150 ml
ka= 3.50 * 10^-4
1 answer
1 answer