demander
- 21
- 0
Hi to all, hope you can help me with a problem that took me almost all the week
To find The Pka of Neutral Red, i had to use this expression \frac{A-A_{HNR}}{A_{NR}-A}
So Now I Have to show that \frac{[NR]}{[HNR^{+}]}=\frac{A-A_{HNR}}{A_{NR}-A}
i tried backwards and i did it, but starting from the concnetrations is far more dificult
i Have to prove that equallity using only this equations
C_{HNR^{+}} + C_{NR}=C_{Total}(1)
A_{HNR^{+}}=\epsilon_{HNR^{+}}.b.C_{total}(2)
A_{NR}=\epsilon_{NR}.b.C_{total}(3)
A=\epsilon_{HNR^{+}}.b.C_{HNR^{+}} + \epsilon_{NR}.b.C_{NR} (4)
I tried to solve 4 in order to C_{HNR^{+}} first and then in order to C_{NR}.
I arrive to
C_{HNR^{+}}=\frac{A-\epsilon_{NR}.b.C_{NR}}{\epsilon_{HNR^{+}}.b}
C_{NR}=\frac{A-\epsilon_{HNR^{+}}.b.C_{HNR^{+}}}{\epsilon_{NR}.b}
then i did the following, added and subtracted the same value in the fraction numerator, like this:
C_{NR}=\frac{A+\epsilon_{HNR^{+}}.b.C_{NR}-\epsilon_{HNR^{+}}.b.C_{HNR^{+}}-\epsilon_{HNR^{+}}.b.C_{NR}}{\epsilon_{NR}.b}
so i could do:
C_{NR}=\frac{A+\epsilon_{HNR^{+}}.b.C_{NR}-\epsilon_{HNR^{+}}.b.(C_{HNR^{+}}+C_{NR}}{\epsilon_{NR}.b}
and using equation 1 it came
C_{NR}=\frac{A+\epsilon_{HNR^{+}}.b.C_{NR}-\epsilon_{HNR^{+}}.b.(C_{t})}{\epsilon_{NR}.b}
then using equation 2:
C_{NR}=\frac{A+\epsilon_{HNR^{+}}.b.C_{NR}-A_{HNR^{+}}}{\epsilon_{NR}.b}
doing the same thing to HNR it came:
C_{HNR}=\frac{A+\epsilon_{NR}.b.C_{HNR}-A_{NR}}{\epsilon_{HNR^{+}}.b}
So doing the reason
\frac{[NR]}{[HNR^{+}]}=\frac{\frac{A+\epsilon_{HNR^{+}}.b.C_{NR}-A_{HNR^{+}}}{\epsilon_{NR}.b}}{\frac{A+\epsilon_{NR}.b.C_{HNR}-A_{NR}}{\epsilon_{HNR^{+}}.b}}
and it's here where i can't see how can i arrive to the final equation
am i going for the worst way? i thought about this all week and can't find a way to prove what i should
hope someone can help me, it's kind of urgent
Homework Statement
To find The Pka of Neutral Red, i had to use this expression \frac{A-A_{HNR}}{A_{NR}-A}
So Now I Have to show that \frac{[NR]}{[HNR^{+}]}=\frac{A-A_{HNR}}{A_{NR}-A}
i tried backwards and i did it, but starting from the concnetrations is far more dificult
Homework Equations
i Have to prove that equallity using only this equations
C_{HNR^{+}} + C_{NR}=C_{Total}(1)
A_{HNR^{+}}=\epsilon_{HNR^{+}}.b.C_{total}(2)
A_{NR}=\epsilon_{NR}.b.C_{total}(3)
A=\epsilon_{HNR^{+}}.b.C_{HNR^{+}} + \epsilon_{NR}.b.C_{NR} (4)
The Attempt at a Solution
I tried to solve 4 in order to C_{HNR^{+}} first and then in order to C_{NR}.
I arrive to
C_{HNR^{+}}=\frac{A-\epsilon_{NR}.b.C_{NR}}{\epsilon_{HNR^{+}}.b}
C_{NR}=\frac{A-\epsilon_{HNR^{+}}.b.C_{HNR^{+}}}{\epsilon_{NR}.b}
then i did the following, added and subtracted the same value in the fraction numerator, like this:
C_{NR}=\frac{A+\epsilon_{HNR^{+}}.b.C_{NR}-\epsilon_{HNR^{+}}.b.C_{HNR^{+}}-\epsilon_{HNR^{+}}.b.C_{NR}}{\epsilon_{NR}.b}
so i could do:
C_{NR}=\frac{A+\epsilon_{HNR^{+}}.b.C_{NR}-\epsilon_{HNR^{+}}.b.(C_{HNR^{+}}+C_{NR}}{\epsilon_{NR}.b}
and using equation 1 it came
C_{NR}=\frac{A+\epsilon_{HNR^{+}}.b.C_{NR}-\epsilon_{HNR^{+}}.b.(C_{t})}{\epsilon_{NR}.b}
then using equation 2:
C_{NR}=\frac{A+\epsilon_{HNR^{+}}.b.C_{NR}-A_{HNR^{+}}}{\epsilon_{NR}.b}
doing the same thing to HNR it came:
C_{HNR}=\frac{A+\epsilon_{NR}.b.C_{HNR}-A_{NR}}{\epsilon_{HNR^{+}}.b}
So doing the reason
\frac{[NR]}{[HNR^{+}]}=\frac{\frac{A+\epsilon_{HNR^{+}}.b.C_{NR}-A_{HNR^{+}}}{\epsilon_{NR}.b}}{\frac{A+\epsilon_{NR}.b.C_{HNR}-A_{NR}}{\epsilon_{HNR^{+}}.b}}
and it's here where i can't see how can i arrive to the final equation
am i going for the worst way? i thought about this all week and can't find a way to prove what i should

hope someone can help me, it's kind of urgent