91Ó°ÊÓ

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

$$\begin{gathered} \overline{\mathrm{x}}=10.160{\mathrm{μ³¾´Ç\pm ô°ä\pm ô}}^{-} \\ \mathrm{s}=2.707{\mathrm{μ³¾´Ç\pm ô°ä\pm ô}}^{-} \end{gathered}$$

$$\begin{gathered} 95\%\mathrm{confidence}\mathrm{interval}=\overline{\mathrm{x}}Â\pm \frac{\mathrm{ts}}{\sqrt{\mathrm{n}}} \\ 95\%\mathrm{confidence}\mathrm{interval}=10.160Â\pm \frac{2.262×1.707}{\sqrt{10}} \\ 95\%\mathrm{confidence}\mathrm{interval}=10.160Â\pm 1.936\mathrm{μ³¾´Ç\pm ô}\mathrm{of}{\mathrm{Cl}}^{-} \end{gathered}$$

Experiment 2

$$\begin{gathered} \overline{\mathrm{x}}=10.770{\mathrm{μ³¾´Ç\pm ô°ä\pm ô}}^{-} \\ \mathrm{s}=3.205{\mathrm{μ³¾´Ç\pm ô°ä\pm ô}}^{-} \end{gathered}$$

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

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

95% confidence interval $=10.770Â\pm 2.293\mathrm{μ³¾´Ç\pm ô}\mathrm{of}{\mathrm{Cl}}^{-}$

Therefore the confidence interval of

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

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

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 ${\mathrm{SO}}_4^{2-}$

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

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

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

In Experiment 1, the precipitate included additional 10.10$\mathrm{μ³¾´Ç\pm ô}$of ${\mathrm{Cl}}^{-}$which equals to $5.08\mathrm{μ³¾´Ç\pm ô}$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