Question

If 35.0 mL of a 0.0500 M HNO3, 35.0 mL of a 0.0600 M KSCN, and 45.0 mL of a 0.0800 M Fe(NO3)3 are combined, what is the initial concentration of SCN- in the mixture?

Answer 1

Step-by-step explanation:

Chemical equation for the given reaction is as follows.

`KSCN \rightarrow K^(+) + SCN^(-)`

KSCN is present 35.0 ml of 0.06 M. Hence, we will calculate the total milimoles of
`SCN^(-)` as follows.

`35.0 * 0.06 M`

= 2.1 mmoles

Now, we will calculate the total volume in mixture as follows.

Total volume = (35 + 35 + 45) mL

= 115 ml

Hence, initial concentration of
`SCN^(-)` will be calculated as follows.

=
`(2.1 mmol)/(115 ml)`

= 0.0182 M

Thus, we can conclude that the initial concentration of
`SCN^(-)` in the mixture is 0.0182 M.