Question

What quantity of 0.25 M HNO3 can be neutralized by 0.10 liters of 0.50 M NaOH?

Answer 1

Answer on Question #42320, Chemistry, Other

Task:

What quantity of $0.25\ \mathrm{M}\ \mathrm{HNO}_3$ can be neutralized by 0.10 liters of $0.50\ \mathrm{M}\ \mathrm{NaOH}$?

Answer:

$$\mathrm{HNO}_3 + \mathrm{NaOH} = \mathrm{NaNO}_3 + \mathrm{H}_2\mathrm{O}$$ $$\bar{N}_{\mathrm{M}} = \frac{v}{V} \quad V = \frac{v}{\bar{N}_{\mathrm{M}}} \quad v = \bar{N}_{\mathrm{M}} \cdot V$$

where $C_{\mathrm{M}}$ – molarity of a solution, M;

$v$ – amount of moles of a substance, moles;

$V$ – volume of a solution, l.

According to the equation, the amount of $\mathrm{HNO}_3$ moles is equal to the amount of NaOH moles.

That is why:

$$C_{\mathrm{M}}(\mathrm{HNO}_3) \cdot V(\mathrm{HNO}_3) = C_{\mathrm{M}}(\mathrm{NaOH}) \cdot V(\mathrm{NaOH})$$ $$V(\mathrm{HNO}_3) = \frac{C_{\mathrm{M}}(\mathrm{NaOH}) \cdot V(\mathrm{NaOH})}{C_{\mathrm{M}}(\mathrm{HNO}_3)}$$ $$V(\mathrm{HNO}_3) = \frac{0.10 \cdot 0.50}{0.25} = 0.20\ \mathrm{l}$$

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