The molarity of KOH is 0.98 M.
Explanation:
KOH + HBr →KBr + H₂O
As per the above reaction, equal moles of KBr and HBr reacts to form 1 mole of KBr and 1 mole of water. We have to find the molarity of KOH by using the law of Volumetric analysis using the equation as,
V1M1 = V2M2
Here V1 and M1 are the volume and molarity of HBr.
V2 and M2 are the volume and molarity of KOH .
We can find the molarity of KOH as,
[tex]$ M2= \frac{V1\times M1}{V2}[/tex]
Plugin the values as,
[tex]$ M2 = \frac{15.40 ml \times 1.4 M}{22.10 ml}[/tex]
= 0.98 M
So the molarity of KOH is 0.98 M.
