Question

If I have 15.40mL of 1.4M HBr(aq) and add 22.10mL KOH what is the molarity of
KOH being used?​

Answer 1

The molarity of KOH is 0.98 M.

Step-by-step 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,

`$ M2= (V1* M1)/(V2)`

Plugin the values as,

`$ M2 = (15.40 ml * 1.4 M)/(22.10 ml)`

= 0.98 M

So the molarity of KOH is 0.98 M.