0
$\begingroup$

Would someone help me solve a task:

The concentration of chloride ions in aqueous solutions can be determined by the Volhardt method. An excess of silver nitrate is added to the sample containing the chlorides, and the extremely insoluble silver chloride is precipitated. Titration determines the excess of silver ions, from which the concentration of chloride ions in the initial sample can be determined. However, silver nitrate is a rather expensive reagent. One alternative that could be used in the same method is some soluble lead(II) salt, eg lead(II) nitrate. However, lead(II) chloride is slightly more soluble than silver chloride, so it is necessary to calculate whether the use of lead(II) salts will affect the precision of the method.

If a solution of lead(II) nitrate (7.45 g in 100 mL of water) is added to 200 mL of a sodium chloride solution with a concentration of 0.1023 mol/L, how much lead(II) chloride will precipitate? What is the concentration of lead(II) ions in the solution after precipitation? What is the equilibrium concentration of chloride ions remaining in the solution after percipitation? Calculate the error in the determination of chloride ions. The solubility product of lead(II) chloride is $1.7 \times 10^{-5}\,\text{mol}^3\,\text{L}^{-3}$.

Solution:

First, I determined the limiting reagent, which in this case is sodium chloride. This helped me to calculate the mass of precipitated lead(II) chloride which is 2.84 g. I then calculated the mass of excess lead(II) ions and determined their concentration in the final solution (300 mL) to be 0.041 mol/L. I'm not entirely sure about the chloride ion concentration calculation -- I calculated it from the expression for the solubility product $$[\ce{Cl^-}] = \sqrt{\frac{K_\text{sp}}{[\ce{Pb^2+}]}}$$ and by including the calculated concentration of lead(II) ions present in the solution I got 0.02039 mol/L. I am struggling with determining the error of the chloride ion concentration since I only calculated one chloride concentration.

I would be very grateful if someone could check $[\ce{Cl^-}]$ and help me calculating the error.

$\endgroup$
3
  • 3
    $\begingroup$ Why do you want to use Pb and introduce an error and work with Pb? You seem to be working with large amounts of reagents! rewrite your method to use 10mL samples, use a simple Mohr titration and a 10mL low actinic buret and use as little silver nitrate as possible. Another consideration is to buy a chloride sensing electrode. [or possibly a conductivity meter]. PbCl2 is too soluble for quant analysis. $\endgroup$
    – jimchmst
    Aug 24, 2023 at 23:46
  • $\begingroup$ crvenikupus - What you are really looking for in this situation is the negative bias in the analysis. You assume that the analysis should yield 100% of the chloride, then you calculate how much chloride is really left in solution. The chloride left in the solution is the negative bias. $\endgroup$
    – MaxW
    Aug 25, 2023 at 4:12
  • $\begingroup$ @MaxW hmm, okay. do you think that I calculated the equilibrium concentration of chloride correctly? $\endgroup$ Aug 25, 2023 at 9:40

1 Answer 1

1
$\begingroup$

Ok, your formula for calculating the chloride concentration:

$$[\ce{Cl^-}] = \sqrt{\frac{K_\text{sp}}{[\ce{Pb^2+}]}}$$

is fine as far as it goes. The equation can give a ballpark estimate of the method's bias. However the solubility product of lead(II) chloride, $1.7×10^{−5}$ is only known to two significant figures, and the solutions are concentrated enough that activities should be used instead of concentrations.

For the lead $$\ce{(7.45 g)/(331.2 g/mol) = 0.0225 mol}$$

For the chloride $$\ce{(0.2 L) x (0.1023 mol/L) = 0.0205 }$$

No since each lead cation precipitates two chloride anions the moles of excess $\ce{Pb}$ is:

$$0.0225 - (0.0205/2) = 0.0123$$

In 300 ml of solution this gives a lead molar concentration of

$$ 0.0123/0.300 = 0.041$$

So: $$[\ce{Cl^-}] = \sqrt{\frac{K_\text{sp}}{[\ce{Pb^2+}]}}=\sqrt{\frac{1.7×10^{−5}}{0.041}} = 0.0204$$

Now for 300 ml of solution the moles of chloride is: $$0.3 * 0.0204 = 0.00611$$

which means that there is a loss of

$$ 0.00611/0.0205 *100\% = 29.8\%$$

of the chloride. So the method with lead will have a very significant negative bias.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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