Recent content by hc772

So if the absorbance of 0.335 is only due to the picric anion the concentration of the picric anion of the solution is as follows: 0.335 = (1.973)CA CA= 0.173 M Ka = [A+]/[NaA][NaOH] = [0.173]/[.12][1] = 1.44 M

the molar absorptivity of the picric acid would be 0.581 = a(0.300 M) a= (0.581/0.300) = 1.937 M and for the mixture of sodium picrate and NaOH it would be 0.335 = a1(0.120 M) + a2(1 M) This gives me two unknowns for one equation. can I use the molar absorptivity of the picric acid...

So to find the equilibrium concentrations I did: NaA + NaOH ---> A + OH + 2Na 0.012 M 1.0 M 0 0 - -x -x +x +x - 0.12 - x 1.0 -x x x - Kc = [A][OH]/[NaH][NaOH] = (0.12 -x)(1.0 -x)/(x)(x)...

I thought I was given the concentrations in the question? will the [OH]- not matter because [H]+ >> [OH]-?

Homework Statement A 0.120 M solution of sodium picrate ina 1 M sodium hydroxide solution was observed to have an absorbance of 0.335, due only to the absorption by the picrate anion. In the same spectrophotometer cell and at the same wavelength as in the previous measurement, a 0.300 M...