College

Consider the neutralization reaction:

\[ 2HNO_3(aq) + Ba(OH)_2(aq) \rightarrow 2H_2O(l) + Ba(NO_3)_2(aq) \]

A 0.105 L sample of an unknown \( HNO_3 \) solution required 46.5 mL of 0.100 M \( Ba(OH)_2 \) for complete neutralization. What is the concentration of the \( HNO_3 \) solution?

Answer :

Answer: The concentration of [tex]HNO_3[/tex] comes out to be 0.088 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 [tex]Ba(OH)_2[/tex]

We are given:

Conversion factor: 1 L = 1000 mL

[tex]n_1=1\\M_1=?M\\V_1=0.105L=105mL\\n_2=2\\M_2=0.100M\\V_2=46.5mL[/tex]

Putting values in above equation, we get:

[tex]1\times M_1\times 105=2\times 0.100\times 46.5\\\\M_1=0.088M[/tex]

Hence, the concentration of [tex]HNO_3[/tex] comes out to be 0.088 M.

Final answer:

The concentration of the HNO3 solution is 0.443 M.

Explanation:

To determine the concentration of the HNO3 solution, we can use the equation, concentration = (volume of solution titrated)/(volume of solution required for complete neutralization). In this case, the volume of solution titrated is 0.105 L and the volume of 0.100 M Ba(OH)2 solution required for complete neutralization is 46.5 mL (or 0.0465 L). Plugging these values into the equation, we get: concentration = 0.0465 L / 0.105 L = 0.443 M.

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