Chemistry, asked by navedansari7653, 10 months ago

You are provided with 90 ml of distilled water and 10 ml of concentrated sulphuric acid to prepare dilute sulphuric acid. (i) what is the correct way of preparing dilute sulphuric acid? Give reason. (ii) how will the concentration of h3o + ions change on dilution?

Answers

Answered by monica789412
9

The correct way is to make dilution with slow adding of acid and continuous stirring.

When acid is made to dilute the H30+ ion concentration decreases in the solution.

Explanation:

The addition of concentrated sulphuric acid in distill water is a highly exothermic process.

Therefore, to get the acid diliuted it should be poured very slowly and stirred continuosly.

The concentration of H3O+ ion decreases on dilution of the acid. When water is added in the acid the H+ ions of the acid reacts with OH- of water, this way H30+ ions of acid decreases.

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Answered by sundarkrishna350
0

Explanation:

In order to dilute an acid, first of all you have to decide the concentration of desired (C2) and of course you have to know the concentration of the original acid solution that you wanna dilute (C1). In addition you have to decide the volume of interest of the diluted solution you gonna get (V2).

Make sure that C1 and C2 are given in the same unit for example both of them in molarity (M).

After you get these values you have to use this formula: C1.V1=C2.V2 which means:

solution's concentration x its volume = the diluted solutions' concentration x its volume

Using this formula you can get the value of V1. Now you know how much voulme you need to take from the original solution. In order to know how much water you should add to V1 calculate it by: V2-V1 the result is the amount of water which you have to add to V1.

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