0
$\begingroup$

The solubility of sodium tetraphenylborate in water is $\pu{470 g/L}$ while the solubility of potassium tetraphenylborate in water is $\pu{1.8e–4 g/L}$. Due to the orders of magnitude difference in aqueous solubility, $\ce{NaB(C6H5)4}$ is used to gravimetrically analyze the potassium content of fertilizer:

$$\ce{K+(aq) + NaB(C6H5)4(aq) -> KB(C6H5)4(s) + Na+(aq)}$$

The water soluble components of a $\pu{10.7875 g}$ sample of 15-25-15 fertilizer were separated from the water insoluble components using vacuum filtration. The filtrate was treated with an excess of $\ce{NaB(C6H5)4}$ and the resulting solid was isolated, dried, and the mass of the solid was determined. Potassium tetraphenylborate $(\pu{12.4393 g})$ was recovered from the gravimetric analysis of $\pu{10.7875 g}$ fertilizer.

What is the percent of potassium as $\ce{K2O}$ in the fertilizer sample?

I know how to get to the answer; I am just unsure how to use the solubility that was given to get there. I also am not sure why the resulting dried mass is higher than the initial mass of fertilizer:

$$\ce{K+(aq) + \underset{\pu{10.7875 g}}{NaB(C6H5)4(aq)} -> \underset{\pu{12.4393 g}}{KB(C6H5)4(s)} + Na+(aq)}$$

$$M(\ce{NaB(C6H5)4}) = \pu{342.22 g mol-1}$$ $$M(\ce{KB(C6H5)4}) = \pu{358.35 g mol-1}$$

$$\pu{12.4393 g~\ce{KB(C6H5)4}}\cdot\frac{\pu{1 mol}}{\pu{342.22 g}}\cdot\frac{\pu{1 mol}~\ce{K}}{\pu{1 mol}~\ce{KB(C6H5)4}}\cdot\frac{\pu{1 mol}~\ce{K2O}}{\pu{2 mol}~\ce{K}}\cdot\frac{\pu{94.2 g}~\ce{K2O}}{\pu{1 mol}~\ce{K2O}} = \pu{1.712 g}~\ce{K2O}$$

$$1~\text{mass} = \frac{\pu{1.71 g}~\ce{K2O}}{\pu{10.7875 g}}\cdot 100\% = 15.9\%$$

$\endgroup$
  • 1
    $\begingroup$ You made a minor calculation mistake in the MW of the salt. It should be 358.35 g/mole not 342.22 g/mol. You want the MW of the potassium salt, not the sodium salt. $\endgroup$ – MaxW Apr 5 '19 at 23:20
4
$\begingroup$

The solubility has been given solely to justify the application of the given method of gravimetric analysis. Since the majority of potassium salts are well or moderately soluble in water, for this analysis it's assumed that entire potassium content is brought to the solution. Since the solubility value of potassium tetraphenylborate is over a million times smaller than the corresponding value for the analyte (sodium tetraphenylborate), it's safe to speak of the completeness of the analysis as all dissolved potassium species were isolated as tetraphenylborate salt and analyzed.

As for the mass difference, that shouldn't be of your concern as the content of the fertilizer isn't given in the first place and you cannot make any assumptions regarding molecular masses of its constituents. Plus, tetraphenylborate $\ce{B(C6H5)4-}$ is a bulky anion of substantial molecular mass $(M(\ce{B(C6H5)4-}) = \pu{319.23 g mol-1})$, which is much higher in comparison to "usual suspects" found in fertilizers, such as phosphates $(M(\ce{PO4^3-}) = \pu{94.97 g mol-1})$ and nitrates $(M(\ce{NO3-}) = \pu{62.00 g mol-1})$.

Long story short, your answer seems correct to me, but I propose to solve the problem algebraically first, starting with what has to be determined and omit intermediate calculations, unless you were explicitly asked to perform any. To illustrate my approach and use of the notations for physical quantities and units, I leave my solution here.

Content of potassium expressed via potassium oxide is a mass fraction $ω(\ce{K2O})$, which is the ratio between the mass of potassium oxide $m(\ce{K2O})$ and the total mass of fertilizer taken for analysis $m_\mathrm{tot}$:

$$ω(\ce{K2O}) = \frac{m(\ce{K2O})}{m_\mathrm{tot}}$$

Unknown mass $m(\ce{K2O})$ can be found from the corresponding amount $n(\ce{K2O})$ and molecular mass $M(\ce{K2O})$:

$$m(\ce{K2O}) = n(\ce{K2O})\cdot M(\ce{K2O})$$

In turn, amount $n(\ce{K2O})$ can be determined from the reaction's stoichiometry:

$$n(\ce{K2O}) = \frac{1}{2}n(\ce{K+}) = \frac{1}{2}n(\ce{KB(C6H5)}) = \frac{m(\ce{KB(C6H5)})}{2\cdot M(\ce{KB(C6H5)})}$$

Now, since we are at the point when no unknowns are left, lets put everything together so that the first equation can be rewritten as such:

$$ \begin{align} ω(\ce{K2O}) &= \frac{m(\ce{KB(C6H5)})\cdot M(\ce{K2O})}{2\cdot m_\mathrm{tot}\cdot M(\ce{KB(C6H5)})} \\ &= \frac{\pu{12.4393 g}\cdot\pu{94.20 g mol-1}}{2\cdot\pu{10.7875 g}\cdot\pu{358.32 g mol-1}} \\ &= \pu{0.152}~\text{or}~15.2\% \end{align} $$

$\endgroup$
  • 2
    $\begingroup$ Think you should clarify this a bit in that the assumption is that all the potassium species are soluble. $\endgroup$ – MaxW Apr 5 '19 at 23:18
  • 2
    $\begingroup$ @MaxW Good point; I left a link to the Wikipedia solubility chart for potassium salts. Thanks! $\endgroup$ – andselisk Apr 5 '19 at 23:30
  • 1
    $\begingroup$ We tend to be sloppy about assumptions on the site. I had a high school teacher who absolutely considered the assumptions as being part of the answer. She was really a fantastic teacher. I took one year of high chemistry under her and exempted two quarters of freshman college chemistry. $\endgroup$ – MaxW Apr 5 '19 at 23:33
  • 1
    $\begingroup$ @Alchimista That's because SE mobile app is handicapped and doesn't support several MathJax features such as \pu{...}, \dfrac, arrays etc. Mobile app is suited for StackOverflow and receiving notifications only. See Undefined control sequence \pu in official SE Android app; MathJax in the Android App (I'm experiencing déjà vu as I feel like you asked me about this a month or so ago :)). $\endgroup$ – andselisk Apr 6 '19 at 8:28
  • 1
    $\begingroup$ :True now I remember. I have to cope with this as for at the moment myt phone is small to open the browser. $\endgroup$ – Alchimista Apr 6 '19 at 8:31

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.