Statistics of coprecipitation. 17In Experiment 1,200.0mL of solution containing 10.0mg of ${\mathbf{SO}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}$(from ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$) were treated with excess ${\mathbf{Bacl}}_{\mathbf{2}}$ solution to precipitate ${\mathbf{BaSO}}_{\mathbf{4}}$ containing some coprecipitated ${\mathbf{Cl}}^{\mathbf{-}}$. To find out how much coprecipitated was present, the precipitate was dissolved in 35mL of 98wt%${\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$ and boiled to liberate , which was removed by bubbling gas through the ${\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$. The HCI/${\mathbf{N}}_{\mathbf{2}}$ stream was passed into a reagent solution that reacted with to give a color that was measured. Ten replicate trials gave values of 7.8,9.8,7.8,7.8,7.8,7.8,13,7,12.7,13.7, and 12.7. Experiment 2 was identical to the first one, except that the mL solution also contained of from ). Ten replicate trials gave 7.8,10,8,8.8,7.8,6.9, 8.8, 15.7 , 12.7 , 13.7and $14.7{\mathrm{\mu molCl}}^{-}$.

(a) Find the mean, standard deviation, and 95% confidence interval for ${\mathbf{Cl}}^{\mathbf{-}}$in each experiment.

(b) Is there a significant difference between the two experiments? What does your answer mean?

(c) If there were no coprecipitate, what mass of ${\mathbf{BaSO}}_{\mathbf{4}}$(FM 233.39) would be expected?

(d) If the coprecipitate is (FM 208.23), what is the average mass of precipitate$\left({\mathrm{BaSO}}_4+{\mathrm{BaCl}}_2\right)$in Experiment 1. By what percentage is the mass greater than the mass in part (c)?

Coprecipitation (CPT) or co-precipitation is the carrying down by a precipitate of substances that are normally soluble under the conditions used in chemistry. Coprecipitation is an important topic in chemical analysis, where it can be both undesirable and useful.

(a)To find the mean and standard deviation

Experiment 1

$$\begin{gathered} \overline{\mathrm{x}}=10.160{\mathrm{\mu molCl}}^{-} \\ \mathrm{s}=2.707{\mathrm{\mu molCl}}^{-} \end{gathered}$$

$$\begin{gathered} 95\%\mathrm{confidence}\mathrm{interval}=\overline{\mathrm{x}}\pm \frac{\mathrm{ts}}{\sqrt{\mathrm{n}}} \\ 95\%\mathrm{confidence}\mathrm{interval}=10.160\pm \frac{2.262\times 1.707}{\sqrt{10}} \\ 95\%\mathrm{confidence}\mathrm{interval}=10.160\pm 1.936\mathrm{\mu mol}\mathrm{of}{\mathrm{Cl}}^{-} \end{gathered}$$

Experiment 2

$$\begin{gathered} \overline{\mathrm{x}}=10.770{\mathrm{\mu molCl}}^{-} \\ \mathrm{s}=3.205{\mathrm{\mu molCl}}^{-} \end{gathered}$$

95% confidence interval$=\overline{\mathrm{x}}\pm \frac{\mathrm{ts}}{\sqrt{\mathrm{n}}}$

95% confidence interval$=10.770\pm \frac{2.262\times 3.205}{\sqrt{10}}$

95% confidence interval $=10.770\pm 2.293\mathrm{\mu mol}\mathrm{of}{\mathrm{Cl}}^{-}$

Therefore the confidence interval of

(b)Yes there is a significant difference between two experiments.

$$\begin{gathered} {\mathrm{S}}_{\mathrm{pooled}}=\sqrt{\frac{{\mathrm{s}}_1^2\left({\mathrm{n}}_1-1\right)+{\mathrm{s}}_2^2\left({\mathrm{n}}_2-1\right)}{{\mathrm{n}}_1+{\mathrm{n}}_2}}=\sqrt{\frac{2.{707}^2\left(10-1\right)+3.{205}^2\left(10-1\right)}{10+10-2}}=2.966 \\ {\mathrm{t}}_{\mathrm{calculated}}=\frac{\left|\overline{\mathrm{x}}-{\overline{\mathrm{x}}}_2\right|}{{\mathrm{S}}_{\mathrm{pooled}}}\sqrt{\frac{{\mathrm{n}}_1{\mathrm{n}}_2}{{\mathrm{n}}_1+{\mathrm{n}}_2}}=\frac{\left|10.160-19.770\right|}{2.966}\sqrt{\frac{10\times 10}{10+10}}=\left|0.61\right|=0.61 \end{gathered}$$

Note that ${\mathrm{t}}_{\mathrm{calculated}}$has a positive value due to$\left|10.160-10.770\right|=\left|-0.61\right|=0.61$

The ${\mathrm{t}}_{\mathrm{calculated}}<{\mathrm{t}}_{\mathrm{tabulated}}$for 18$°$of freedom for $95 \%$ confidence interval - the difference wouldn ' be significant

- our result means that addition of excess ${\mathrm{Cl}}^{-}$before the precipitation wouldn't lead to coprecipitation of${\mathrm{Cl}}^{-}$

Considering that it is given in the task that we have 10mg of ${\mathrm{SO}}_4^{2-}$from ${\mathrm{Na}}_2{\mathrm{SO}}_4$we will use that info in order to find the mass of ${\mathrm{BaSO}}_4$

Finding the moles of ${\mathrm{SO}}_4^{2-}$

$\mathrm{n}\left({\mathrm{SO}}_4^{2-}\right)=\mathrm{m}/\mathrm{M}=\left(10\times {10}^{-3}\mathrm{g}\right)/96.07\mathrm{g}/\mathrm{mol}=1.041\times {10}^{-4}\mathrm{mol}$

(c)Calculating the mass of ${\mathrm{BaSO}}_4$

$\mathrm{m}\left({\mathrm{BaSO}}_4\right)=\mathrm{n}\times \mathrm{M}=\left(1.041\times {10}^{-4}\mathrm{mol}\right)\times 208.23\mathrm{g}/\mathrm{mol}=24.295\mathrm{mg}$

In Experiment 1, the precipitate included additional 10.10$\mathrm{\mu mol}$of ${\mathrm{Cl}}^{-}$which equals to $5.08\mathrm{\mu mol}$of ${\mathrm{BaCl}}_2=1.058\mathrm{mg}$of ${\mathrm{BaCl}}_2$

(d)The average mass of precipitate $\left({\mathrm{BaSO}}_4+{\mathrm{BaCl}}_2\right)$in Experiment 1 is 1.058mg .

An increase in the mass would be $1.058\mathrm{mg}/24.295\mathrm{mg}=4.35\%$and it would represent a significant error for the analysis