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0.450 L of 0.0500 M HCl is titrated to the equivalence point with 21.45 mL of a NaOH solution. What is the concentration (in M) of the NaOH solution that was added?

User Alex Leo
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1 Answer

4 votes

Answer:

1.0489 M is the concentration (in M) of the NaOH solution that was added

Step-by-step explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:


n_1M_1V_1=n_2M_2V_2

where,


n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is
HCl


n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:


n_1=1\\M_1=0.0500 M\\V_1=0.450 L\\n_2=1\\M_2=?\\V_2=21.45 mL=0.02145 L

Putting values in above equation, we get:


1* 0.0500 M* 0.450 L=1* M_2* 0.02145 L\\\\M_2=1.0489 M

1.0489 M is the concentration (in M) of the NaOH solution that was added

User Hubi
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8.3k points
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