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

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

Answer 2

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 $KOH$.

$1\times M_1\times 90ml=2\times 1.50M\times 25.7ml$

$M_1=0.856M$

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