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2 years ago

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]2 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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<h3>Answer:</h3>

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<h3>General Formulas and Concepts:</h3>

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The metals will lose electrons while the non metals will gain electrons in order to attain octet structure.

An ion can be cation (positively charged) or anion (negatively charged).

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O ------------ 6 ---------------------- 2 ------------------------ none -------------- $O^{2-}$

Ca -------- 2 ----------------------- none ---------------------- 2 ------------------ $Ca^{2+}$

Br ----------- 7 --------------------- 1 ------------------------ none --------------- $Br^{-}$

S ------------ 6 ----------------------- 2 ------------------------ none --------------- $S^{2-}$

Cl ------------ 7 ----------------------- 1 ------------------------ none ---------------- $Cl^{-}$

K -------------- 1 ----------------------- none ----------------------- 1 ------------------ $K^{+}$

Mg ------------ 2 ---------------------- none ---------------------- 2 ---------------- $Mg^{2+}$

Be ------------- 2 ---------------------- none ---------------------- 2 ---------------- $Be^{2+}$

Learn more here: brainly.com/question/21089350

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2 years ago

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

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