Question

Consider a solution containing 0.100 m fluoride ions and 0.126 m hydrogen fluoride. the concentration of hydrogen fluoride after addition of 5.00 ml of 0.0100 m hcl to 25.0 ml of this solution is __________ m.

Answer 1

The concentration of hydrogen fluoride after addition of 5.00 mL of 0.0100 M HCl to 25.0 mL of solution of fluoride ions and hydrogen fluoride is $\boxed{0.107{\text{ M}}}$.

Further Explanation:

Different concentration terms are utilized in determining the concentration of various solutions. Some of the most commonly used terms are written below.

1. Molarity (M)

2. Mole fraction (X)

3. Molality (m)

4. Parts per million (ppm)

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

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

Molarity is one of the concentration terms used very often in solutions. It is defined as moles of solute present in one litre of solution. The formula to calculate molarity of solution is as follows:

${\text{Molarity of solution}} = \dfrac{{{\text{Moles }}\left( {{\text{mol}}} \right){\text{of solute}}}}{{{\text{Volume }}\left( {\text{L}} \right){\text{ of solution}}}}$                                     …… (1)

Rearrange equation (1) to calculate moles of solute.

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

Substitute 0.0100 M for molarity of solution and 5.0 mL for volume of solution in equation (2) to calculate moles of HCl.

$\begin{aligned}{\text{Moles of HCl}} &= \left( {0.0100{\text{ M}}} \right)\left( {5.0{\text{ mL}}}\right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{{\text{1 mL}}}}} \right) \\&= 0.00005{\text{ mol}} \\\end{aligned}$  

Substitute 0.126 M for molarity of solution and 25.0 mL for volume of solution in equation (2) to calculate moles of HF.

$\begin{aligned}{\text{Moles of HF}} &= \left( {0.126{\text{ M}}} \right)\left( {25.0{\text{ mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{{\text{1 mL}}}}} \right) \\&= 0.00315{\text{ mol}} \\\end{aligned}$  

Number of moles of HF after addition of HCl can be calculated as follows:

$\begin{aligned}{\text{Moles of HF}} &= 0.00315{\text{ mol}} + 0.00005{\text{ mol}} \\&= {\text{0}}{\text{.0032 mol}} \\\end{galigned}$  

The total volume of solution is calculated as follows:

$\begin{aligned}{\text{Total volume of solution}} &= {\text{5}}{\text{.0 mL}} + 25.0{\text{ mL}} \\&={\text{30}}{\text{.0 mL}} \\\end{aligned}$  

Substitute 0.0032 mol for the moles of solute and 30.0 mL for volume of solution in equation (1) to calculate the concentration or molarity of HF solution.

$\begin{aligned}{\text{Molarity of solution}} &= \left( {\frac{{{\text{0}}{\text{.0032 mol}}}}{{{\text{30}}{\text{.0 mL}}}}} \right)\left( {\frac{{1{\text{ mL}}}}{{{{10}^{ - 3}}{\text{ L}}}}} \right) \\&= 0.107{\text{ M}} \\\end{aligned}$  

Learn more:

1. Calculation of volume of gas: brainly.com/question/3636135
2. Determine how many moles of water produce: brainly.com/question/1405182

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Concentration terms

Keywords: concentration, concentration terms, HF, HCl, 0.107 M, 25.0 mL, molarity, moles, volume, 30.0 mL, 5.0 mL, 0.126 M, 0.0032 mol, 0.00005 mol, 0.00315 mol.

Answer 2

Fluorine ions reacts with Hydrogen chloride to form more hydrogen fluoride.
Therefore, moles of HCl = 0.005 l × 0.01 M = 5 ×10^-5 moles
The initial moles of Hydrogen fluoride will be;
 = 0.0126 M× 0.0250 = 0.00315 Moles
Moles of hydrogen fluoride after the addition of HCl 
= 0.00315 + 5.0× 10^-5 = 0.0032 moles 
Therefore, the concentration of Hydrogen chloride 
            = 0.0032 moles/ 0.030 L 
            = 0.107 M