Question

What volume of 5.00×10−3mhno3 is needed to titrate 20.00 ml of 5.00×10−3mca(oh)2 to the equivalence point?

Answer 1

The balanced equation for the given titration reaction can be written as

$2 HNO_{3}(aq)+ Ca(OH)_{2}(aq)\rightarrow Ca(NO_{3})_{2}+ 2 H_{2}O$

Step 1 : Find moles of Ca(OH)₂

Volume of Ca(OH)₂ = 20 mL

Concentration of Ca(OH)₂ = 5.00 x 10⁻³ M

Moles of Ca(OH)₂ = $20 mL * \frac{1L}{1000 mL} * \frac{5.00*10^{-3}mol}{L}$

Moles of Ca(OH)₂ = 1.00 x 10⁻⁴

Step 2 : Find moles of HNO3 using mole ratio from balanced equation

The mole ratio of HNO₃ and Ca(OH)₂ is 2 : 1

Let us use this as a conversion factor to find moles of HNO₃.

Moles of HNO₃ = $1 * 10^{-4} mol (Ca(OH)_{2} ) * \frac{2mol(HNO_{3})}{1mol(Ca(OH)_{2})}$

Moles of HNO₃ = 2.00 x 10⁻⁴

Step 3 : Find concentration of HNO3

Concentration of HNO3 can be calculated as

$[HNO_{3} ] = \frac{mol (HNO_{3})}{Volume (L)}$

The concentration of HNO₃ is given as 5.00 x 10⁻³

Moles of HNO3 = 2.00 x 10⁻⁴

Let us plug in these values in above equation.

$5 * 10^{-3} = \frac{2.00*10^{-4}}{Volume (L)}$

$Volume = \frac{2.00*10^{-4}}{5*10^{-3}}$

$Volume = 0.04L$

Volume of HNO₃ in mL = $0.04 L * \frac{1000 mL}{1L} = 40 mL$

Volume of HNO₃ needed for titration is 40 mL

Answer 2

$\boxed{{\text{40 mL}}}$of $5\times{10^{-3}}\;{\text{M}}$ ${\text{HN}}{{\text{O}}_{\text{3}}}$ is needed to titrate 20 mL of $5\times{10^{-3}}\;{\text{M}}$ ${\text{Ca}}{\left({{\text{OH}}} \right)_{\text{2}}}$ to the equivalence point.

Further Explanation:

Molarity is a concentration term that is defined as the number of moles of solute dissolved in one liter of the solution. It is denoted by M and its unit is mol/L.

The formula to calculate the molarity of the solution is as follows:

${\text{Molarity of solution}}=\frac{{{\text{Amount}}\left({{\text{mol}}}\right){\text{of solute}}}}{{\;{\text{volume}}\left({\text{L}}\right)\;{\text{of solution}}}}$          ...... (1)

Rearrange equation (1) to calculate the amount of solute.

${\text{Amount of solute}}=\left({{\text{Molarity of solution}}}\right)\left({{\text{Volume}}\;{\text{of}}\;{\text{solution}}}\right)$       …… (2)

Firstly, the volume of   is to be converted into L. The conversion factor for this is,

${\text{1 mL}}={10^{-3}}\;{\text{L}}$

So the volume of ${\text{Ca}}{\left({{\text{OH}}}\right)_{\text{2}}}$ is calculated as follows:

$\begin{gathered}{\text{Volume of Ca}}{\left({{\text{OH}}}\right)_{\text{2}}}=\left({{\text{20 mL}}}\right)\left({\frac{{{{10}^{-3}}\;{\text{L}}}}{{{\text{1}}\;{\text{mL}}}}}\right)\\={\text{2}}\times{\text{1}}{{\text{0}}^{-2}}{\text{ L}}\\\end{gathered}$

The molarity of ${\text{Ca}}{\left({{\text{OH}}}\right)_{\text{2}}}$ the solution is $5\times{10^{-3}}\;{\text{M}}$.

The volume of ${\text{Ca}}{\left({{\text{OH}}}\right)_{\text{2}}}$ the solution is ${\text{2}}\times{\text{1}}{{\text{0}}^{-4}}{\text{ L}}$.

Substitute these values in equation (2) to calculate the amount of ${\text{Ca}}{\left({{\text{OH}}}\right)_{\text{2}}}$.

$\begin{gathered}{\text{Amount of Ca}}{\left({{\text{OH}}}\right)_{\text{2}}}=\left({{\text{5}}\times{\text{1}}{{\text{0}}^{-3}}\;{\text{M}}}\right)\left({{\text{2}}\times{\text{1}}{{\text{0}}^{-2}}\;{\text{L}}}\right)\\=1\times{10^{-4}}\;{\text{mol}}\\\end{gathered}$

The chemical equation for the reaction of nitric acid with calcium hydroxide is as follows:

${\text{Ca}}{\left({{\text{OH}}}\right)_2}+2{\text{HN}}{{\text{O}}_3}\to{\text{Ca}}{\left({{\text{N}}{{\text{O}}_{\text{3}}}}\right)_2}+{{\text{H}}_{\text{2}}}{\text{O}}$

According to the balanced chemical reaction, one mole of ${\text{Ca}}{\left({{\text{OH}}}\right)_{\text{2}}}$ reacts with two moles of ${\text{HN}}{{\text{O}}_3}$ to form products.

So the amount of ${\text{HN}}{{\text{O}}_{\text{3}}}$is twice the amount of ${\text{Ca}}{\left({{\text{OH}}}\right)_{\text{2}}}$and calculated as follows:

${\text{Amount of HN}}{{\text{O}}_{\text{3}}}={\text{2}}\left({{\text{Amount of Ca}}{{\left({{\text{OH}}}\right)}_{\text{2}}}}\right)$              …… (3)

Substitute $1\times{\text{1}}{{\text{0}}^{-4}}{\text{ mol}}$ for the amount of ${\text{Ca}}{\left({{\text{OH}}}\right)_{\text{2}}}$ in equation (3).

$\begin{gathered}{\text{Amount of HN}}{{\text{O}}_{\text{3}}}={\text{2}}\left({{\text{1}}\times{\text{1}}{{\text{0}}^{-4}}\;{\text{mol}}}\right)\\={\text{2}}\times{\text{1}}{{\text{0}}^{-4}}\;{\text{mol}}\\\end{gathered}$

Rearrange equation (1) to calculate the volume of the solution as follows:

${\text{Volume}}\;{\text{of}}\;{\text{solution}}=\frac{{{\text{Amount of solute}}}}{{{\text{Molarity of solution}}}}$            …… (4)

Substitute $2\times{\text{1}}{{\text{0}}^{-4}}{\text{ mol}}$ for the amount of solute and $5\times{10^{-3}}\;{\text{M}}$ for the molarity of solution in equation (4) to calculate the volume of ${\text{HN}}{{\text{O}}_{\text{3}}}$.

$\begin{gathered}{\text{Volume}}\;{\text{of}}\;{\text{HN}}{{\text{O}}_3}=\frac{{{\text{2}}\times{\text{1}}{{\text{0}}^{-4}}\;{\text{mol}}}}{{{\text{5}}\times{\text{1}}{{\text{0}}^{-3}}\;{\text{M}}}}\\={\text{0}}{\text{.04 L}}\\\end{gathered}$

The volume is to be converted into mL. The conversion factor for this is,

${\text{1 L}}={10^3}\;{\text{mL}}$

So the volume of ${\text{HN}}{{\text{O}}_{\text{3}}}$ can be calculated as follows:

$\begin{gathered}{\text{Volume of HN}}{{\text{O}}_{\text{3}}}=\left({{\text{0}}{\text{.04 L}}}\right)\left({\frac{{{{10}^3}\;{\text{mL}}}}{{{\text{1 L}}}}}\right)\\={\text{40 mL}}\\\end{gathered}$

So the volume of ${\mathbf{HN}}{{\mathbf{O}}_{\mathbf{3}}}$ is 40 mL.

Learn more:

1. Determine the given mass of solute: https://brainly.com/question/2847466

2. Determine the moles of water produced: https://brainly.com/question/1405182

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Solutions

Keywords: KOH, molarity, 20 mL, 0.20 L, concentration, concentration terms, volume, moles, molar mass, 0.04 L, 40 mL.