Question

After standardizing a NaOH solution, you use it to titrate an HCl solution known to have a concentration of 0.205 M. You perform five titrations and obtain the following results: 0.213, 0.204, 0.208, 0.200, and 0.198 M.
a) What is the mean?
b) What is the standard deviation of mean?

Answer 1

Answer:

The right answer is:

(a) 0.205

(b) 0.005425

Step-by-step explanation:

The given data is:

0.213, 0.204, 0.208, 0.200, and 0.198

(a)

The mean will be:

= $\frac{Sum \ of \ all \ data}{No. \ of \ data}$

= $\frac{0.213+ 0.204+ 0.208+ 0.200+0.198}{5}$

= $\frac{1.023}{5}$

= $0.2046$

or,

= $0.205$

(b)

The standard deviation will be:

$\sigma^2=\Sigma\frac{(x_i-\mu)^2}{N}$

    $=\frac{(0.213-0.2046)^2+...+(0.198-0.2046)^2}{5}$

    $=\frac{0.0001472}{5}$

    $=2.994\times 10^{-5}$

or,

    $=0.005425$