How many nitrate ions are present in the following aqueous solution?
5.08 L of a solution containing 2.68 x 1021 formula units of lithium nitrate per liter.
When you have the number, determine its log (base 10) and enter that value with 3 decimal places

Respuesta :

Answer:

22.1

Explanation:

Step 1: Calculate the number of formula units of lithium nitrate

A 5.08 L solution contains 2.68 × 10²¹ formula units per liter.

5.08 L × 2.68 × 10²¹ formula units/1 L = 1.36 × 10²² formula units

Step 2: Calculate the number of nitrate ions

Lithium nitrate dissociates completely in water according to the following equation.

LiNO₃(aq) ⇒ Li⁺(aq) + NO₃⁻(aq)

The molar ratio of LiNO₃ to NO₃⁻ is 1:1. The number of nitrate ions is 1/1 × 1.36 × 10²² = 1.36 × 10²².

Then,

log (1.36 × 10²²) = 22.1

The number of nitrate ions are present in the following aqueous solution is 13.6x10²¹, and log(base 10) value of this is 22.2.

What are strong electrolytes?

Those salts which are completely dissociate into their ions in the solution form will known as strong electrolyte.

Lithium nitrate is a strong electrolyte and it shows complete dissociation as:

LiNO₃ → Li⁺ + NO₃⁻

Given that,

1 liter of solution containing = 2.68 x 10²¹ formula unit

5.08 L of solution containing = 5.08 x 2.68 x 10²¹ = 13.6x10²¹ formula unit

So, number of nitrate ion in given solution = 13.6x10²¹ formula unit

Log(13.6x10²¹) = 22.1

Hence, required values are 13.6x10²¹ and 22.1.

To know more about strong electrolytes, visit the below link:
https://brainly.com/question/5547666

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