What are the two species in highest concentration in a 0.1 M solution of lactic acid (CH3CH(OH)C00H), (Ka

1.4 × 10^-4.

Explanation:

C3H6O3 + H2O <> C3H5O3^- + H3O^+ (1).

So, from the question above we are given the following parameters or data which is going to help in solving this particular Question/problem;

=>concentration of the solution of lactic acid (CH3CH(OH)C00H) = 0.1 M and pH = 2.44.

Therefore, the concentration of the hydrogen ion[H^+} can be determined from the pH formula given below;

pH = - log { H^+}.

2.44 = - log { H^+}.

Therefore, {H^+} = 0.0036 M.

From the equation (1) given above, we have that the ratio for the equilibrium reaction is 1 : 1 : 1 :1. Therefore, molarity of C3H5O3^- = 0.0036 M and the molarity of C3H6O3 =( 0.1 - 0.0036 M) = 0.0964 M at equilibrium.

Hence, ka = {C3H5O3^-} { H3O^+} /{C3H6O3} = ( 0.0036 M)^2 /(0.0964 M) = 1.4 × 10^-4.

43.452

step-by-step explanation:

284 times .153 = 43.452

your answer would be “dry”