a) Given the replicate chloride determinations (ppm): 35.8, 36.4, 36.1, 36.0, 36.7. Compute the mean…

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a) Given the replicate chloride determinations (ppm): 35.8, 36.4, 36.1, 36.0, 36.7. Compute the mean, Variance and sample standard deviation (9 marks) b) A burette is calibrated and the readings are n

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QUESTION TWO

a) Given the replicate chloride determinations (ppm): 35.8, 36.4, 36.1, 36.0, 36.7. Compute the mean, Variance and sample standard deviation (9 marks)

Given data points ( $x_i$ ): 35.8, 36.4, 36.1, 36.0, 36.7
Number of data points ( $n$ ) = 5

Calculate the mean ( $\bar{x}$ ).
$\bar{x} = \frac{\sum x_i}{n} = \frac{35.8 + 36.4 + 36.1 + 36.0 + 36.7}{5} = \frac{181.0}{5} = 36.2$
The mean is $\boxed{\text{36.2 ppm}}$ .
Calculate the variance ( $s^2$ ).
First, calculate the sum of squared differences from the mean:
$\sum (x_i - \bar{x})^2 = (35.8 - 36.2)^2 + (36.4 - 36.2)^2 + (36.1 - 36.2)^2 + (36.0 - 36.2)^2 + (36.7 - 36.2)^2$ $= (-0.4)^2 + (0.2)^2 + (-0.1)^2 + (-0.2)^2 + (0.5)^2$ $= 0.16 + 0.04 + 0.01 + 0.04 + 0.25 = 0.50$
Now, calculate the sample variance:
$s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} = \frac{0.50}{5-1} = \frac{0.50}{4} = 0.125$
The variance is $\boxed{\text{0.125 ppm}^2}$ .
Calculate the sample standard deviation ( $s$ ).
$s = \sqrt{s^2} = \sqrt{0.125} \approx 0.35355$
The sample standard deviation is $\boxed{\text{0.354 ppm}}$ .

b) A burette is calibrated and the readings are normally distributed with standard deviation $\sigma = 0.020 \text{ ml}$ . For a titration whose average volume $V = 24.86 \text{ ml}$ for 3 titrations, give a 95% confidence interval for the true of the average volume (6 marks)

Given:
Population standard deviation ( $\sigma$ ) = $0.020 \text{ ml}$
Sample mean ( $\bar{x}$ ) = $24.86 \text{ ml}$
Number of titrations ( $n$ ) = 3
Confidence level = 95%
Determine the critical Z-value for a 95% confidence interval.
For a 95% confidence level, the Z-value is 1.96.
Calculate the margin of error (ME).
$ME = Z \times \frac{\sigma}{\sqrt{n}} = 1.96 \times \frac{0.020 \text{ ml}}{\sqrt{3}}$ $ME = 1.96 \times \frac{0.020 \text{ ml}}{1.732}$ $ME = 1.96 \times 0.011547 \text{ ml} \approx 0.02263 \text{ ml}$
Calculate the 95% confidence interval.
$\text{Confidence Interval} = \bar{x} \pm ME$ $\text{Confidence Interval} = 24.86 \text{ ml} \pm 0.02263 \text{ ml}$ $\text{Lower bound} = 24.86 - 0.02263 = 24.83737 \text{ ml}$ $\text{Upper bound} = 24.86 + 0.02263 = 24.88263 \text{ ml}$
The 95% confidence interval for the true average volume is $\boxed{\text{24.837 ml to 24.883 ml}}$ .

c) Formic acid is a weak acid with a dissociation constant Ka of $1.8 \times 10^{-4}$ . The Ka relates the concentration of H $^+$ ions and amount of acid dissolved, M by the equation: $K_a = \frac{[H^+]^2}{M-[H^+]}$ . Given that there is 0.1 moles of formic acid dissolved, calculate the pH of the solution. (pH = -log[H $^+$ ]) (5 marks)

Given:
$K_a = 1.8 \times 10^{-4}$
Initial concentration of formic acid ( $M$ ) = $0.1 \text{ M}$ (assuming 1 L solution for 0.1 moles)
Equation: $K_a = \frac{[H^+]^2}{M-[H^+]}$
Substitute the given values into the $K_a$ expression.
$1.8 \times 10^{-4} = \frac{[H^+]^2}{0.1 - [H^+]}$
Rearrange the equation into a quadratic form.
Let $x = [H^+]$ .
$1.8 \times 10^{-4} (0.1 - x) = x^2$ $1.8 \times 10^{-5} - 1.8 \times 10^{-4} x = x^2$ $x^2 + 1.8 \times 10^{-4} x - 1.8 \times 10^{-5} = 0$
Solve the quadratic equation for $x$ (which is $[H^+]$ ) using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .
Here, $a=1$ , $b=1.8 \times 10^{-4}$ , $c=-1.8 \times 10^{-5}$ .
$x = \frac{-(1.8 \times 10^{-4}) \pm \sqrt{(1.8 \times 10^{-4})^2 - 4(1)(-1.8 \times 10^{-5})}}{2(1)}$ $x = \frac{-1.8 \times 10^{-4} \pm \sqrt{3.24 \times 10^{-8} + 7.2 \times 10^{-5}}}{2}$ $x = \frac{-1.8 \times 10^{-4} \pm \sqrt{0.0000720324}}{2}$ $x = \frac{-1.8 \times 10^{-4} \pm 0.008487}{2}$
Since $[H^+]$ cannot be negative, we take the positive root:
$x = [H^+] = \frac{-1.8 \times 10^{-4} + 0.008487}{2} = \frac{0.008307}{2} = 0.0041535 \text{ M}$
Calculate the pH of the solution.
$\text{pH} = -\log[H^+] = -\log(0.0041535)$ $\text{pH} \approx 2.381$
The pH of the solution is $\boxed{\text{2.38}}$ .

A) Given the replicate chloride determinations (ppm): 35.8, 36.4, 36.1, 36.0, 36.7. Compute the mean, Variance and sample standard deviation (9 marks) b) A burette is calibrated and the readings are n

Step-by-step solution

Calculate the mean ( $\bar{x}$ ).

Calculate the variance ( $s^2$ ).

Calculate the sample standard deviation ( $s$ ).

Determine the critical Z-value for a 95% confidence interval.

Calculate the margin of error (ME).

Calculate the 95% confidence interval.

Substitute the given values into the $K_a$ expression.

Rearrange the equation into a quadratic form.

Solve the quadratic equation for $x$ (which is $[H^+]$ ) using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .

Calculate the pH of the solution.

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