Answer:- [tex]\frac{[In^-]}{[HIn]}=3.16*10^-^8[/tex]
Solution:- If both acid and basic form are present at the same time then it's like a buffer solution since the buffer solution also has a weak acid and its conjugate base. The Handerson equation is used for solving buffer problems. Let's use the same equation here also.
The equation is:
[tex]pH=pKa+log(\frac{base}{acid})[/tex]
pKa is given as 9.7 and the pH is 2.2. let's plug in the values in the equation.
[tex]2.2=9.7+log(\frac{[In^-]}{[HIn]})[/tex]
subtract 9.7 from sides:
[tex]2.2-9.7=log(\frac{[In^-]}{[HIn]})[/tex]
[tex]-7.5=log(\frac{[In^-]}{[HIn]})[/tex]
taking antilog:
[tex]10^-^7^.^5=\frac{[In^-]}{[HIn]}[/tex]
[tex]3.16*10^-^8=\frac{[In^-]}{[HIn]}[/tex]
So, the answer is [tex]3.16*10^-^8[/tex] .
