Chapter 27, Problem 27.38P

(a)

Interpretation Introduction

Interpretation:

For ${\text{Cl}}^{-}$, mean, standard deviation and 95% confidence interval in each experiments has to be found.

Concept introduction:

Mean or average $\overline{(x})$: Add up all given numbers and divided by number of number.

Standard deviation (s):

$\text{s = }\sqrt{\frac{{\displaystyle \sum _{\text{i =1}}^{\text{N}}{\left({\text{x}}_{\text{i}}\text{-}\overline{\text{x}}\right)}^{\text{2}}}}{\text{N-1}}}$

Confidence interval: It can take any number of possibilities, with most common being 95% to 99%.

Answer to Problem 27.38P

Experiment: 1

$\overline{\text{x }}\text{=10}\text{.160 μmol}$

$\text{s =2}\text{.707 μmol}$

95% confidence interval = $\text{10}\text{.160}\pm \text{1}\text{.936 μmol}$

Experiment: 2

$\overline{\text{x }}\text{=10}\text{.770 μmol}$

$\text{s =3}\text{.205 μmol}$

95% confidence interval = $\text{10}\text{.770}\pm 2.293\text{ μmol}$

Explanation of Solution

For experiment: 1

To find: mean for ${\text{Cl}}^{-}$

$\begin{array}{l}\overline{\text{x}}\text{ =}\frac{\text{7}\text{.8+9}\text{.8+7}\text{.8+7}\text{.8+7}\text{.8+7}\text{.8+13}\text{.7+12}\text{.7+13}\text{.7+12}\text{.7}}{\text{10}}\\ \\ \text{=}\text{10}\text{.160 μmol}\end{array}$

To find: standard deviation for ${\text{Cl}}^{-}$

1. 1- Calculate the mean or average of given values.
2. 2- For each number, subtract the average. Square the result
3. 3- Adding all of the squared result
4. 4- Divided this sum by (N-1)

Finally obtain the standard deviation for Chlorine is $\text{s =2}\text{.707 μmol}$.

To find: 95% confidence interval for ${\text{Cl}}^{-}$

$\begin{array}{l}\text{95\% confidence interval = }\overline{\text{x}}\text{±}\frac{\text{ts}}{\sqrt{\text{n}}}\\ \\ \text{=10}\text{.160±}\frac{\left(\text{2}\text{.262}\right)\left(\text{2}\text{.707}\right)}{\sqrt{\text{10}}}\\ \\ \text{=10}\text{.160±1}\text{.936 μmol}{\text{Cl}}^{-}\end{array}$

For experiment: 2

To find: mean for ${\text{Cl}}^{-}$

$\begin{array}{l}\overline{\text{x}}\text{ =}\frac{\text{7}\text{.8+10}\text{.8+8}\text{.8+7}\text{.8+6}\text{.9+8 }\text{.8+15}\text{.7+12}\text{.7+13}\text{.7+14}\text{.7}}{\text{10}}\\ \\ \text{=}\text{10}\text{.770 μmol}\end{array}$

To find: standard deviation for ${\text{Cl}}^{-}$

1. 1- Calculate the mean or average of given values.
2. 2- For each number, subtract the average. Square the result
3. 3- Adding all of the squared result
4. 4- Divided this sum by (N-1)

Finally obtain the standard deviation for Chlorine is $\text{s =3}\text{.205 μmol}$.

To find: 95% confidence interval for ${\text{Cl}}^{-}$

$\begin{array}{l}\text{95\% confidence interval = }\overline{\text{x}}\text{±}\frac{\text{ts}}{\sqrt{\text{n}}}\\ \\ \text{=10}\text{.160±}\frac{\left(\text{2}\text{.262}\right)\left(3.205\right)}{\sqrt{\text{10}}}\\ \\ \text{=10}\text{.770±2}\text{.293 μmol}{\text{Cl}}^{-}\end{array}$

(b)

Interpretation Introduction

Interpretation:

There is any significant difference between given two experiments has to be predicted.

Concept introduction:

Mean or average $\overline{(x})$: Add up all given numbers and divided by number of number.

Answer to Problem 27.38P

There is no significant difference between two experiments.

Explanation of Solution

To explain: any significant difference between two experiments

$\begin{array}{l}{\text{s}}_{\text{pooled}}\text{=}\sqrt{\frac{{\text{s1}}^{\text{2}}\left({\text{n}}_{\text{1}}\text{-1}\right){\text{+ s2}}^{\text{2}}\left({\text{n}}_{\text{2}}\text{-1}\right)}{{\text{n}}_{\text{1}}{\text{+n}}_{\text{2}}\text{-2}}}\\ \\ \text{=}\sqrt{\frac{\text{2}\text{.7072}\left(\text{10-1}\right)\text{+3}\text{.2052}\left(\text{10-1}\right)}{\text{10+10-2}}}\\ \\ \text{=}\text{2}\text{.966}\end{array}$

$\begin{array}{l}{\text{t}}_{\text{calculated}}\text{=}\frac{\left|\overline{\text{x}}\text{1-}\overline{\text{x}}\text{2}\right|}{{\text{s}}_{\text{pooled}}}\sqrt{\frac{{\text{n}}_{\text{1}}{\text{n}}_{\text{2}}}{{\text{n}}_{\text{1}}{\text{+n}}_{\text{2}}}}\\ \\ \text{=}\frac{\left|\text{10}\text{.160-10}\text{.770}\right|}{\text{2}\text{.966}}\sqrt{\frac{\left(\text{10}\right)\left(\text{10}\right)}{\text{10+10}}}\\ \\ \text{= 0}\text{.460 <}{\text{t}}_{\text{tabulated}}\text{for degrees of freedom}\text{for 95\% confidence level}\text{.}\\ \end{array}$

So, there is no difference in significant.  The result means addition of extra ${\text{Cl}}^{\text{-}}$ preceding precipitation.  It does not lead to extra co-precipitation of chlorine under these conditions.    But in general under another conditions would expect additional chlorine to lead to extra co-precipitation.

(c)

Interpretation Introduction

Interpretation:

Expected mass of Barium sulphate has to be calculated if there is no coprecipitation.

Answer to Problem 27.38P

Mass of Barium sulphate is found to be $\text{24}\text{.295mg}$.

Explanation of Solution

To determine: Mass of Barium sulphate

$\begin{array}{l}\text{10}{\text{.0mg of SO}}_{\text{4}}^{\text{2-}}\text{= 0}\text{.10410mmol}\\ \\ \text{=}\text{24}\text{.295mg}{\text{BaSO}}_{\text{4}}\end{array}$

(d)

Interpretation Introduction

Interpretation:

Average mass of precipitate in first experiment has to be calculated and the percentage is the mass greater than mass in part (c) has to be predicted.

Answer to Problem 27.38P

Average mass of Barium sulphate is found to be $\text{1}{\text{.058 mg BaCl}}_2$.

Explanation of Solution

To determine: average mass of Barium chloride

In experiment 1, the precipitate include an extra $\text{10}\text{.160 μmol}{\text{Cl}}^{-}$. So,

$\begin{array}{l}\text{10}{\text{.160 μmol Cl}}^{-}\text{=5}\text{.08}{\text{mg BaCl}}_{\text{2}}\text{=1}{\text{.058mg BaCl}}_{\text{2}}\\ \end{array}$

The increased mass is,

$\begin{array}{l}\left(1.058\right)\left(24.295\right)\text{=}\text{4}\text{.35\%}\\ \end{array}$

This is a large error in this experiment.