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if it takes 54 mL of 0.1 NaOH to neutralize 125 mL of an HCL solution, what is the concentration of HCL?

User Fbielejec
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1 Answer

4 votes

Answer:

0.0432M

Step-by-step explanation:

We begin by writing a balanced equation for the reaction. This is illustrated below:

NaOH + HCl —> NaCl + H2O

From the equation above,

The number of mole of the acid (nA) = 1

The number of mole of the base (nB) = 1

Data obtained from the question include:

Vb (volume of the base) = 54mL

Cb (concentration of the base) = 0.1M

Va (volume of the acid) = 125mL

Ca ( concentration of the acid) =?

Using CaVa/CbVb = nA/nB, the concentration of the acid can easily be obtained as shown below:

CaVa/CbVb = nA/nB

Ca x 125 / 0.1 x 54 = 1

Cross multiply to express in linear form:

Ca x 125 = 0.1 x 54

Divide both side by 125

Ca = (0.1 x 54) / 125

Ca = 0.0432M

Therefore, the concentration of the acid is 0.0432M

User Vasanti
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