A 50.0 ML Solution Of $Ca(OH)_2$ With An Unknown Concentration Was Titrated With 0.340 M $HNO_3$. To Reach The Endpoint, A Total Of 17.2 ML Of $HNO_3$ Was Required.Write The Balanced Chemical Equation Based On The Following

A 50.0 mL Solution of $Ca(OH)_2$ with an Unknown Concentration: Titration with $HNO_3$

In chemistry, titration is a widely used analytical technique for determining the concentration of a substance in a solution. It involves the reaction of a known amount of a substance, called the titrant, with a sample of the unknown substance. The endpoint of the titration is the point at which the reaction is complete, and it is typically indicated by a color change or a pH change. In this article, we will discuss the titration of a 50.0 mL solution of $Ca(OH)_2$ with an unknown concentration with 0.340 M $HNO_3$.

To write the balanced chemical equation for the titration of $Ca(OH)_2$ with $HNO_3$, we need to consider the chemical reaction that occurs between the two substances. The reaction is a neutralization reaction, in which the hydroxide ions from the $Ca(OH)_2$ react with the hydrogen ions from the $HNO_3$ to form water and a salt.

The balanced chemical equation for the reaction is:

$$Ca(OH)_2 + 2HNO_3 \rightarrow Ca(NO_3)_2 + 2H_2O$$

In the titration reaction, the $HNO_3$ is added to the $Ca(OH)_2$ solution until the endpoint is reached. At the endpoint, the reaction is complete, and all of the hydroxide ions from the $Ca(OH)_2$ have reacted with the hydrogen ions from the $HNO_3$ to form water and a salt.

The titration reaction can be represented by the following equation:

$$Ca(OH)_2 + 2H^+ \rightarrow Ca^{2+} + 2H_2O$$

The stoichiometry of the reaction is the ratio of the number of moles of one substance to the number of moles of another substance. In this case, the stoichiometry of the reaction is 1:2, meaning that one mole of $Ca(OH)_2$ reacts with two moles of $HNO_3$.

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To determine the concentration of the $Ca(OH)_2$ solution, we need to use the titration data. The titration data includes the volume of $HNO_3$ required to reach the endpoint, the concentration of the $HNO_3$ solution, and the volume of the $Ca(OH)_2$ solution.

The concentration of the $Ca(OH)_2$ solution can be calculated using the following equation:

$$[Ca(OH)_2] = \frac{n_{Ca(OH)_2}}{V_{Ca(OH)_2}}$$

where $[Ca(OH)_2]$ is the concentration of the $Ca(OH)_2$ solution, $n_{Ca(OH)_2}$ is the number of moles of $Ca(OH)_2$, and $V_{Ca(OH)_2}$ is the volume of the $Ca(OH)_2$ solution.

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To calculate the concentration of the $Ca(OH)_2$ solution, we need to use the titration data. The titration data includes the volume of $HNO_3$ required to reach the endpoint, the concentration of the $HNO_3$ solution, and the volume of the $Ca(OH)_2$ solution.

The volume of $HNO_3$ required to reach the endpoint is 17.2 mL, and the concentration of the $HNO_3$ solution is 0.340 M. The volume of the $Ca(OH)_2$ solution is 50.0 mL.

The number of moles of $HNO_3$ required to reach the endpoint can be calculated using the following equation:

$$n_{HNO_3} = C_{HNO_3} \times V_{HNO_3}$$

where $n_{HNO_3}$ is the number of moles of $HNO_3$, $C_{HNO_3}$ is the concentration of the $HNO_3$ solution, and $V_{HNO_3}$ is the volume of the $HNO_3$ solution.

The number of moles of $HNO_3$ required to reach the endpoint is:

$$n_{HNO_3} = 0.340 \, M \times 0.0172 \, L = 0.00586 \, mol$$

The number of moles of $Ca(OH)_2$ can be calculated using the stoichiometry of the reaction. The stoichiometry of the reaction is 1:2, meaning that one mole of $Ca(OH)_2$ reacts with two moles of $HNO_3$.

The number of moles of $Ca(OH)_2$ is:

$$n_{Ca(OH)_2} = \frac{n_{HNO_3}}{2} = \frac{0.00586 \, mol}{2} = 0.00293 \, mol$$

The concentration of the $Ca(OH)_2$ solution can be calculated using the following equation:

$$[Ca(OH)_2] = \frac{n_{Ca(OH)_2}}{V_{Ca(OH)_2}}$$

where $[Ca(OH)_2]$ is the concentration of the $Ca(OH)_2$ solution, $n_{Ca(OH)_2}$ is the number of moles of $Ca(OH)_2$, and $V_{Ca(OH)_2}$ is the volume of the $Ca(OH)_2$ solution.

The concentration of the $Ca(OH)_2$ solution is:

$$[Ca(OH)_2] = \frac{0.00293 \, mol}{0.0500 \, L} = 0.0586 \, M$$

In this article, we discussed the titration of a 50.0 mL solution of $Ca(OH)_2$ with an unknown concentration with 0.340 M $HNO_3$. We wrote the balanced chemical equation for the reaction, and we calculated the concentration of the $Ca(OH)_2$ solution using the titration data. The concentration of the $Ca(OH)_2$ solution was found to be 0.0586 M.

• CRC Handbook of Chemistry and Physics, 97th ed., CRC Press, 2016.
• Chemical Equilibrium, 2nd ed., McGraw-Hill, 2013.
• Analytical Chemistry, 7th ed., Wiley, 2012.
  A 50.0 mL Solution of $Ca(OH)_2$ with an Unknown Concentration: Titration with $HNO_3$ - Q&A

In our previous article, we discussed the titration of a 50.0 mL solution of $Ca(OH)_2$ with an unknown concentration with 0.340 M $HNO_3$. We wrote the balanced chemical equation for the reaction and calculated the concentration of the $Ca(OH)_2$ solution using the titration data. In this article, we will answer some frequently asked questions about the titration of $Ca(OH)_2$ with $HNO_3$.

A: The purpose of titration is to determine the concentration of a substance in a solution. Titration involves the reaction of a known amount of a substance, called the titrant, with a sample of the unknown substance.

A: A strong acid is an acid that completely dissociates in water to produce hydrogen ions, whereas a weak acid is an acid that only partially dissociates in water to produce hydrogen ions.

A: A strong base is a base that completely dissociates in water to produce hydroxide ions, whereas a weak base is a base that only partially dissociates in water to produce hydroxide ions.

A: The endpoint in titration is the point at which the reaction is complete, and it is typically indicated by a color change or a pH change.

A: The concentration of a substance is determined in titration by using the titration data, including the volume of the titrant required to reach the endpoint, the concentration of the titrant, and the volume of the sample.

A: The titrant is the substance that is added to the sample in known amounts until the endpoint is reached. The titrant is typically a strong acid or a strong base.

A: The stoichiometry of the reaction in titration is the ratio of the number of moles of one substance to the number of moles of another substance. The stoichiometry of the reaction is used to calculate the concentration of the substance in the sample.

A: The concentration of a substance is calculated in titration by using the titration data, including the volume of the titrant required to reach the endpoint, the concentration of the titrant, and the volume of the sample. The concentration of the substance is calculated using the following equation:

$$[Ca(OH)_2] = \frac{n_{Ca(OH)_2}}{V_{Ca(OH)_2}}$$

where $[Ca(OH)_2]$ is the concentration of the $Ca(OH)_2$ solution, $n_{Ca(OH)_2}$ is the number of moles of $Ca(OH)_2$, and $V_{Ca(OH)_2}$ is the volume of the $Ca(OH)_2$ solution.

In this article, we answered some frequently asked questions about the titration of $Ca(OH)_2$ with $HNO_3$. We discussed the purpose of titration, the difference between strong and weak acids and bases, the significance of the endpoint in titration, and the role of the titrant in titration. We also discussed the significance of the stoichiometry of the reaction in titration and how the concentration of a substance is calculated in titration.

• CRC Handbook of Chemistry and Physics, 97th ed., CRC Press, 2016.
• Chemical Equilibrium, 2nd ed., McGraw-Hill, 2013.
• Analytical Chemistry, 7th ed., Wiley, 2012.