The Ka for benzoic acid is 6.5 × 10−5. Calculate the pH of a 0.14 M benzoic acid solution.

2.52

Explanation:

C6H5COOH + H2O ⇄ C6H5COO- + H30+

 0.14M                                0                   0

  -X                                     +X                  +X

---------------------------------------------------------------

0.14-X                                   X                    X

Ka= 6.5 × 10^-5 =  

Since the effect of X in equation (0.14-x) is very small, it can be ignored.

x= 3.02 × 10^-3

-logx = 2.52

answer: the figure is attached herewith.

step-by-step explanation:   we are given to sketch a unit circle and label the terminal points corresponding to

in the attached figure, we have drawn a unit circle, with centre at o(0,0) and radius 1 unit. we have labelled the points a, b, c, d and e corresponding to the angles below

∠aox = 60°,

∠box = 120°,

∠cox = 180°,

∠dox = 240°,

∠eox = 300°.

and the coordinates are a(0.5,1), b(-0.5,1), c(-1,0), d(-0.5,-1) and e(0.5,-1).

thus, the answer is complete.

i believe it is a.) released into

explanation: