Chemistry, 14.07.2019 22:00, yolomcswaggin20

25.0 ml of an hbr solution were titrated with 29.15 ml of a 0.205 m lioh solution to reach the equivalence point. what is the molarity of the hbr solution?

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answered: amyeileen

HBr reacts with LiOH and forms LiBr and H₂O as the products. The balanced reaction is

LiOH(aq) + HBr(aq) → LiBr(aq) + H₂O(l)

Molarity (M) = moles of solute (mol) / volume of the solution (L)

Molarity of LiOH = 0.205 M

Volume of LiOH = 29.15 mL = 29.15 x 10⁻³ L

Hence,

moles of LiOH = molarity x volume of the solution

= 0.205 M x 29.15 x 10⁻³ L

= 5.97575 x 10⁻³ mol

The stoichiometric ratio between LiOH and HBr is 1 : 1.

Hence,

moles of HBr in 25.0 mL = moles of LiOH added

= 5.97575 x 10⁻³ mol

Hence, molarity of HBr = 5.97575 x 10⁻³ mol / 25.00 x 10⁻³ L

= 0.23903 M

≈ 0.239 M

Hence, the molarity of the HBr is 0.239 M.

answered: mrstealyogirl40

The molarity of the HBr solution is $\boxed{{\text{0}}{\text{.239 M}}}$ .

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, HBr is titrated with LiOH. So molarity equation becomes,

${{\text{M}}_{{\text{HBr}}}}{{\text{V}}_{{\text{HBr}}}} = {{\text{M}}_{{\text{LiOH}}}}{{\text{V}}_{{\text{LiOH}}}}$ ......(1)

Here,

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

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

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

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

Rearrange equation (1) to calculate ${{\text{M}}_{{\text{HBr}}}}$ .

${{\text{M}}_{{\text{HBr}}}}=\dfrac{{{{\text{M}}_{{\text{LiOH}}}}{{\text{V}}_{{\text{LiOH}}}}}}{{{{\text{V}}_{{\text{HBr}}}}}}$ ......(2)

The value of ${{\text{V}}_{{\text{HBr}}}}$ is 25 mL.

The value of ${{\text{M}}_{{\text{LiOH}}}}$ is 0.205 M.

The value of ${{\text{V}}_{{\text{LiOH}}}}$ is 29.15 mL.

Substitute these values in equation (2).

$\begin{aligned}{{\text{M}}_{{\text{HBr}}}} &= \frac{{\left({{\text{0}}{\text{.205 M}}}\right) \times\left( {{\text{29}}{\text{.5 mL}}} \right)}}{{\left( {{\text{25 mL}}} \right)}}\\&= 0.23903{\text{ M}}\\&\approx {\text{0}}{\text{.239 M}}\\\end{aligned}$