Question

A volume of 90.0 ml of aqueous potassium hydroxide (koh) was titrated against a standard solution of sulfuric acid (h2so4). what was the molarity of the koh solution if 25.7 ml of 1.50 m h2so4 was needed? the equation is 2koh(aq)+h2so4(aq)→k2so4(aq)+2h2o(l)

Answer 1

Answer : The molarity of the KOH solution is, 0.856 M

Explanation :

The balanced chemical reaction is,

`2KOH(aq)+H_2SO_4(aq)\rightarrow K_2SO_4(aq)+2H_2O(l)`

Using dilution law,

`n_1M_1V_1=n_2M_2V_2`

where,

`n_1` = acidity of an base = 1

`n_2` = basicity of an acid = 2

`M_1` = concentration of
`KOH` = ?

`M_2` = concentration of
`H_2SO_4` = 1.50 M

`V_1` = volume of
`KOH` = 90 ml

`V_2` = volume of
`H_2SO_4` = 25.7 ml

Now put all the given values in the above law, we get the concentration of the
`KOH`.

`1* M_1* 90ml=2* 1.50M* 25.7ml`

`M_1=0.856M`

Therefore, the molarity of the KOH solution is, 0.856 M

Answer 2

the balanced reaction for the above acid base reaction is as follows;
2KOH + H₂SO₄ ---> K₂SO₄ + 2H₂O
Stoichiometry of KOH to H₂SO₄ is 2:1
The number of HCl moles reacted - 1.50 M / 1000 mL/L x 25.7 mL = 0.0386 mol
Then the number of KOH moles reacted - 0.0386 mol x 2 = 0.0772 mol
The number of moles in 90.0 mL - 0.0772 mol
Therefore in 1000.0 mL - 0.0772 mol / 90.0 mL x 1000 mL = 0.86 mol
therefore molarity of KOH - 0.86 M