Question

Suppose 13.00 mL of 0.100 M barium hydroxide is required to neutralize 17.00 ml of nitric acid with an unknown concentration. What is the concentration of the nitric acid. Express your answer in mol/L

Answer 1

To solve for the concentration of nitric acid in mol/L or also M. We use the equation, 
                                          M₁V₁ = M₂V₂
Where M stands for the molarity and V is for volume. Substituting the known values,
                               (0.10 M)(13.0 mL) = (M₂)(17 mL)
                                                 M₂ = 0.076 M
Thus, the concentration of the nitric acid is approximately 0.076 M. 

Answer 2

The concentration of ${\text{Ba}}{\left( {{\text{OH}}} \right)_2}$solution is $\boxed{{\text{0}}{\text{.076 M}}}$

Further Explanation:

The concentration is the proportion of substance in the mixture. The most commonly used concentration terms are as follows:

1. Molarity (M)

2. Molality (m)

3. Mole fraction (X)

4. Parts per million (ppm)

5. Mass percent ((w/w) %)

6. Volume percent ((v/v) %)

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

The molarity equation is given by the following expression:

${{\text{M}}_{\text{1}}}{{\text{V}}_{\text{1}}} = {{\text{M}}_{\text{2}}}{{\text{V}}_{\text{2}}}$

Here,

${{\text{M}}_{\text{1}}}$ is the molarity of the first solution.

${{\text{V}}_{\text{1}}}$ is the volume of the first solution.

${{\text{M}}_{\text{2}}}$ is the molarity of the second solution.

${{\text{V}}_{_{\text{2}}}}$ is the volume of the second solution.

In the concerned question, ${\text{Ba}}{\left( {{\text{OH}}} \right)_2}$is neutralized by ${\text{HN}}{{\text{O}}_{\text{3}}}$. So molarity equation becomes,

${{\text{M}}_{{\text{Ba}}{{\left( {{\text{OH}}} \right)}_2}}}{{\text{V}}_{{\text{Ba}}{{\left( {{\text{OH}}} \right)}_2}}} = {{\text{M}}_{{\text{HN}}{{\text{O}}_{\text{3}}}}}{{\text{V}}_{{\text{HN}}{{\text{O}}_{\text{3}}}}}$                           …… (1)

Here,

${{\text{M}}_{{\text{Ba}}{{\left( {{\text{OH}}} \right)}_2}}}$ is the molarity of ${\text{Ba}}{\left( {{\text{OH}}} \right)_2}$ solution.

${{\text{V}}_{{\text{Ba}}{{\left( {{\text{OH}}} \right)}_2}}}$is the volume of ${\text{Ba}}{\left( {{\text{OH}}} \right)_2}$solution.

${{\text{M}}_{{\text{HN}}{{\text{O}}_{\text{3}}}}}$ is the molarity of ${\text{HN}}{{\text{O}}_{\text{3}}}$solution.

${{\text{V}}_{{\text{HN}}{{\text{O}}_{\text{3}}}}}$ is the volume of ${\text{HN}}{{\text{O}}_{\text{3}}}$solution.

Rearrange equation (1) to calculate ${{\text{M}}_{{\text{HN}}{{\text{O}}_{\text{3}}}}}$.

${{\text{M}}_{{\text{HN}}{{\text{O}}_{\text{3}}}}} = \frac{{{{\text{M}}_{{\text{Ba}}{{\left( {{\text{OH}}} \right)}_2}}}{{\text{V}}_{{\text{Ba}}{{\left( {{\text{OH}}} \right)}_2}}}}}{{{{\text{V}}_{{\text{HN}}{{\text{O}}_{\text{3}}}}}}}$                                       …… (2)

The value of ${{\text{V}}_{{\text{Ba}}{{\left( {{\text{OH}}} \right)}_2}}}$ is 13 mL.

The value of ${{\text{M}}_{{\text{Ba}}{{\left( {{\text{OH}}} \right)}_2}}}$ is 0.100 M.

The value of ${{\text{V}}_{{\text{HN}}{{\text{O}}_{\text{3}}}}}$ is 17 mL.

Substitute these values in equation (2).

$\begin{aligned}{{\text{M}}_{{\text{HN}}{{\text{O}}_{\text{3}}}}}&=\frac{{\left( {{\text{0}}{\text{.100 M}}} \right)\left( {{\text{13 mL}}} \right)}}{{\left( {{\text{17 mL}}}\right)}}\\&=0.{\text{07647 M}}\\&\approx {\text{0}}{\text{.076 M}}\\\end{aligned}$

Hence the concentration of barium hydroxide solution is 0.076 M.

Learn more:

1. What is the mass of 1 mole of viruses: brainly.com/question/8353774

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

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Concentration terms

Keywords: molarity, 13 mL, 17 mL, 0.076 M, Ba(OH)2, M1, V1, M2, V2, HNO3, 0.100 M, molarity, volume.