Find the ph of a 0.250 m solution of nac2h3o2. (the ka value of hc2h3o2 is 1.80×10−5).

When NaC2H3O2 is in solution, it would act as a base since it ionizes into sodium ions and acetate ions, the acetate ions would interact with water with the following reaction:

C2H3O2- + H2O <---> C2H3O2H + OH-  

So, from the reaction, hydroxide ions are formed giving it basic properties. We calculate as follows:

[NaC2H3O2] = [Na+] = [C2H3O2-] = 0.250 m

By the ICE table:
            [C2H3O2-]         C2H3O2H      OH-
I              0.250                     0                0
C                -x                      +x               +x

E            0.250 - x                x                 x         

 1.80×10^−5 = x^2 / (0.250 - x)

Solving for x, we have 

x = 0.002112

pOH = -log 0.002112

pOH = 2.68

pH = 14 - 2.68 = 11.32

Therefore, the pH of the solution is about 11.32.

hey there!

boiling point is the temperature at which the vapor pressure of the liquid equals atmospheric pressure. the vapor pressure of solvent molecules is decreased when a solute is added, so a higher temperature is required to increase the number of solvent molecules in the gas phase above the liquid.

at the freezing point, the vapor pressures of the solid and liquid are equal; a lower temperature is needed to reduce the number of solvent particles above the liquid.

hope this you

tobey