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

The right answer is:

(a) 0.205

(b) 0.005425

Explanation:

The given data is:

0.213, 0.204, 0.208, 0.200, and 0.198

(a)

The mean will be:

=
`(Sum \ of \ all \ data)/(No. \ of \ data)`

=
`(0.213+ 0.204+ 0.208+ 0.200+0.198)/(5)`

=
`(1.023)/(5)`

=
`0.2046`

or,

=
`0.205`

(b)

The standard deviation will be:

`\sigma^2=\Sigma((x_i-\mu)^2)/(N)`

`=((0.213-0.2046)^2+...+(0.198-0.2046)^2)/(5)`

`=(0.0001472)/(5)`

`=2.994* 10^(-5)`

or,

`=0.005425`