Question

A solution containing 80 grams of sodium hydroxide (molecular mass = 40) completely neutralizes another solution containing 98 grams of an acid a (molecular mass = 98). The charge on the anion part of a is

Answer 1

Answer : The charge on the anion part of 'a' is, (-2)

Explanation :

First we have to calculate the moles of sodium hydroxide and an acid 'a'.

$\text{Moles of NaOH}=\frac{\text{Mass of NaOH}}{\text{Molar mass of NaOH}}=\frac{80g}{40g/mole}=2mole$

$\text{Moles of acid}=\frac{\text{Mass of acid}}{\text{Molar mass of acid}}=\frac{98g}{98g/mole}=1mole$

Now we have to determine the charge on the cation part of acid 'a'.

As, 2 moles of sodium hydroxide dissociate to gives 2 moles of sodium ion and 2 moles of hydroxide ion.

So, for the complete neutralization 2 moles of hydroxide ion will neutralizes 2 moles of hydrogen ion.

That means the charge on the cation part of an acid that is hydrogen ion will be, (+2)

Now we have to determine the charge of anion part of an acid 'a'.

According to the electroneutrality principle, each atom in a stable substance has a charge close to zero.

That means, for a stable compound (acid) we require (-2) charge on anion that means to neutralize the (+2) charge of cation of an acid.

Therefore, the charge on the anion part of 'a' is, (-2)

Answer 2

Answer:

A solution containing 80 grams of sodium hydroxide (molecular mass = 40) completely neutralizes another solution containing 98 grams of an acid a (molecular mass = 98). The charge on the anion part of a is -2

Explanation:

The reaction between an acid and a base is called as neutralization reaction. The neutralization reaction between NaOH (molecular mass = 40) and H2SO4 (molecular mass = 98) will be as mentioned below:

H2SO4+2NaOH→Na2SO4+2H2O

From this reaction, we can say that 1 mole of H2SO4 is required to neutralize 2 moles of NaOH.

H2SO4 will be completely dissociated in NaOH during neutralization reaction. H2SO4 can be ionized as H+ and SO4^-2.