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This problem is providing us with the volume of nitric acid that is titrated with 0.18 L of 0.1-M sodium hydroxide and asks for the concentration of the acid. At the end, the result turns out to be 0.045M, according to the following.

Acid-base titrations:

In chemistry, acid-base titrations allow us to quantify the volume or concentration of an acid or base via the following equation:

[tex]M_AV_A=M_BV_B[/tex]

Where the subscript A stands for the acid and B for the base; which means one can calculate any of the variables there by knowing the other three. This equation is based on the balanced neutralization chemical equation, which takes place between the acid and the base.

Thus, we can write the reaction between NaOH and HNO3 as:

[tex]HNO_3+NaOH\rightarrow NaNO_3+H_2O[/tex]

In such a way, we can solve for the concentration of the acid as shown below:

[tex]M_A=\frac{M_BV_B}{V_A} \\\\M_A=\frac{0.1M*0.18L}{0.40L} \\\\M_A=0.045M[/tex]

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