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

Chemistry

1 answer:

postnew [5]3 years ago

8 0

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,

$$ M2= \frac{V1\times M1}{V2}$

Plugin the values as,

$$ M2 = \frac{15.40 ml \times 1.4 M}{22.10 ml}$

= 0.98 M

So the molarity of KOH is 0.98 M.

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