0
$\begingroup$

If I measure the emission spectra of potassium using the following solutions

1) $0.1~\mathrm{M}$ of $\ce{KNO3}$

2) $0.1~\mathrm{M}$ of $\ce{K2SO4}$

will I see any significant differences?

I think there will be no significant differences because the spectrum does not depend on whether in 1) I am using $\ce{K}$ while in 2) I am using $\ce{K2}$ (the emission spectrum should be independent). Am I right?

$\endgroup$
  • 3
    $\begingroup$ You have neither $\ce{K}$ nor $\ce{K2}$. In both cases, you have $\ce{K+}$ ions, and they are the same. $\endgroup$ – Zhe Nov 24 '16 at 4:13
  • $\begingroup$ They look the same but intensities would be different $\endgroup$ – user1420303 Nov 24 '16 at 11:19
1
$\begingroup$

$0.1~\mathrm{M}$ $\ce{KNO3}$ dissociates to give $0.1~\mathrm{M}$ $\ce{K+}$ ions.

$0.1~\mathrm{M}$ $\ce{K2SO4}$ dissociates to give $2 \times 0.1~\mathrm{M}$ $\ce{K+}$ ions.

Hence, in terms of emission intensity, $\ce{K2SO4}$ should give twice the signal if other parameters are same and if the detection response is linear.


Very sensitive instrument might detect the effect of counter-ion but that will be very difficult since the cation itself is always surrounded by water.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.