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?

  • 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

$0.1~\mathrm{M}$ $\ce{KNO3}$ 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.

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