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A 31.5mL aliquot of HNO3 (aq) of unknown concentration wastitrated with 0.0134 M NaOH (aq). It took 23.9 mL of the base toreach the endpoint of the titration. The concentration (M) of theacid was __________.I forget how to do titration problems and cannot seem to find anysimilar examples in my text. Is there a specific formula you usefor these?

Respuesta :

Answer : The concentration (M) of the acid was, 0.0102 M

Explanation :

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

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HNO_3[/tex]

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

We are given:

[tex]n_1=1\\M_1=?\\V_1=31.5mL\\n_2=1\\M_2=0.0134M\\V_2=23.9mL[/tex]

Putting values in above equation, we get:

[tex]1\times M_1\times 31.5mL=1\times 0.0134M\times 23.9mL\\\\M_1=0.0102M[/tex]

Thus, the concentration (M) of the acid was, 0.0102 M

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