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stepan [7]
3 years ago
13

A 35.0 ml sample of 0.225 m hbr was titrated with 42.3 ml of koh. What is the concentration of the koh?

Chemistry
1 answer:
Lemur [1.5K]3 years ago
5 0

Answer:

The concentration of KOH is 0.186 M

Explanation:

First things first, we need too write out the balanced equation between HBr and KOH.

This is given as;

KOH (aq) + HBr (aq) → KBr (aq) + H2O (l)

From the reaction above, we can tell that it takes 1 mole of KOH to react with 1 mole of HBr.

We use the acid base formular in calculating unknown concentrations. This is given as;

\frac{CaVa}{CbVb}  = \frac{na}{nb}

where;

Ca = Concentration of acid

Va = Volume of acid

Cb = Concentration of base

Vb = Volume of base

na = Number of moles of acid

nb = Number of moles of base

KOH is the base and HBr is acid.

Hence;

Ca = 0.225

Va = 35

Cb = ?

Vb = 42.3

na = 1

nb = 1

Making Cb subject of formular we have;

Cb = \frac{CaVaNb}{VbNa}

Cb = (0.225 * 35 * 1) / (42.3 * 1)

Cb = 0.186 M

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Answer:

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Explanation:

Hello,

In this case, during titration at the equivalence point, we find that the moles of the base equals the moles of the acid:

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