Avolume of 90.0 ml of a 0.550 m hno3 solution is titrated with 0.890 m koh. calculate the volume of koh required to reach the equivalence point.

55.61 mL

Explanation:

Given;

Volume of HNO₃ solution, V₁ = 90.0 mL

Molarity of the solution, M₁ = 0.550 M

Molarity of KOH, M₂ = 0.890 M

the reaction between the two is given as:

HNO₃ + KOH ⇒ KNO₃ + H₂O

Here,

1 mol of HNO₃ reacts with 1 mol of KOH

thus, we have

M₁V₁ = M₂V₂

here,

V₂ is the volume of KOH

on substituting the values, we get

0.550 × 90.0 = 0.890 × V₂

or

V₂ = 55.61 mL

Hence,

The volume of KOH required to reach the equivalence point = 55.61 mL

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !