Chemistry, 22.12.2019 03:31 yatayjenings12

In a 1.0× 10–6 m solution of ba(oh)2(aq) at 25 °c, identify the relative molar amounts of these species.

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In a 1.0× 10–6 M solution of $Ba(OH)^{2}$ (aq) at 25 °C, the relative molar amounts of these species are $H_{2}O$ (the most), $OH^{-}$ , $Ba^{2+}$ , $H_{3}O^{+}$ , $Ba(OH)^{2}$ (the least)

Further explanation

Molar concentration, also called molarity, amount concentration or substance concentration, is a measure of the concentration of a chemical species in particular of a solute in a solution in terms of amount of substance per unit volume of solution

$H_{3}O^{+}$ and $NO^{3-}$ are the same at 1.0*10^-6 M because $HNO_{3}$ completely dissociates into equal parts

$[OH^{-}] = \frac{1.0*10^{-14}}{1.0*10^{-6}} = 1.0*10^{-8}$ M

$HNO_{3}$ should not be any if it all dissociates . It is strong acid so 100% dissociation so we don't need to find the molarity of water because it is the solvent

Therefore the answer is:

$H_{2}O$ is the most high relative molar amounts $OH^{-}$ next ( $2.0*10^{-6}$ M) $Ba^{2+}$ next ( $1.0*10^{-6}$ M) (strong base, 100% dissociation) $H_{3}O^{+}$ $= \frac{1.0*10^{-14}}{2.0*10^{-6}} = 5*10^{-9} M$ Ba(OH)2 is the least relative molar amounts (almost none)Learn moreLearn more about the relative molar amounts Learn more about molarity Learn more about Ba(OH)2(aq) Answer details

Grade: 9

Subject: Chemistry

Chapter: the relative molar amounts of species.

Keywords: the relative molar amounts, molar amounts, molar, molarity, Ba(OH)2(aq)

Answer from: marquezsturgis

The amounts of various species in a solution of $1.0 \times {10^{ - 6}}{\text{ M Ba}}{\left( {{\text{OH}}} \right)_2}$ are as follows:

$\boxed{\begin{aligned} \left[ {{\text{B}}{{\text{a}}^{2 + }}} \right] &= 1.0 \times {10^{ - 6}}{\text{ M}} \hfill \\ \left[ {{\text{O}}{{\text{H}}^ - }} \right] &= 2.0 \times {10^{ - 6}}{\text{ M}} \hfill \\ \left[ {{\text{Ba}}{{\left( {{\text{OH}}} \right)}_2}} \right] &= 1.0 \times {10^{ - 6}}{\text{ M}} \hfill \\ \left[ {{{\text{H}}_{\text{3}}}{{\text{O}}^ + }} \right] &= 5.0 \times {10^{ - 9}}{\text{ M}} \hfill \\ \end{aligned}}$

Further Explanation:

Concentration terms are used to describe concentrations of solutions. Some of these are as follows:

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) %)

7. Parts per billion (ppb)

${\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ is a strong base whose dissociation occurs as follows:

${\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}} \rightleftharpoons {\text{B}}{{\text{a}}^{2 + }} + 2{\text{O}}{{\text{H}}^ - }$

This indicates one mole of ${\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ dissociates to produce one mole of ions and two moles of ${\text{O}}{{\text{H}}^ - }$ ions.

Given information:

${\text{Concentration of Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}} = 1.0 \times {10^{ - 6}}{\text{ M}}$

Since one mole of ${\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ forms one mole of ${\text{B}}{{\text{a}}^{2 + }}$ ions, concentration of ${\text{B}}{{\text{a}}^{2 + }}$ ions is same as that of ${\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ and can be written as follows:

${\text{Concentration of B}}{{\text{a}}^{2 + }} = 1.0 \times {10^{ - 6}}{\text{ M}}$

Since one mole of ${\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ forms two moles of ${\text{O}}{{\text{H}}^ - }$ ions, concentration of ions becomes twice the concentration of ${\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ and is calculated as follows:

$\begin{aligned} {\text{Concentration of O}}{{\text{H}}^ - } &= 2\left( {1.0 \times {{10}^{ - 6}}{\text{ M}}} \right) \\ &= 2.0 \times {10^{ - 6}}{\text{ M}} \\ \end{aligned}$

Therefore concentration of ${\text{O}}{{\text{H}}^ - }$ ions comes out to be $2.0 \times {10^{ - 6}}{\text{ M}}$ .

The expression for ${{\text{k}}_{\text{w}}}$ of water is as follows:

${{\text{k}}_{\text{w}}} = \left[ {{{\text{H}}_{\text{3}}}{{\text{O}}^ + }} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]$ …… (1)

Here,

${{\text{k}}_{\text{w}}}$ is ionic product constant of water.

$\left[ {{{\text{H}}_{\text{3}}}{{\text{O}}^ + }} \right]$ is the concentration of hydronium ions.

$\left[ {{\text{O}}{{\text{H}}^ - }} \right]$ is the concentration of hydroxide ions.

Rearrange equation (1) to calculate concentration of ${{\text{H}}_{\text{3}}}{{\text{O}}^ + }$ .

$\left[ {{{\text{H}}_{\text{3}}}{{\text{O}}^ + }} \right] = \dfrac{{{{\text{k}}_{\text{w}}}}}{{\left[ {{\text{O}}{{\text{H}}^ - }} \right]}}$ …… (2)

Substitute ${10^{ - 14}}$ for ${{\text{k}}_{\text{w}}}$ and $2.0 \times {10^{ - 6}}{\text{ M}}$ for $\left[ {{\text{O}}{{\text{H}}^ - }} \right]$ in equation (2).

$\begin{aligned} \left[ {{{\text{H}}_{\text{3}}}{{\text{O}}^ + }} \right] &= \frac{{{{10}^{ - 14}}}}{{2.0 \times {{10}^{ - 6}}{\text{ M}}}} \\ &= 5.0 \times {10^{ - 9}}{\text{ M}} \\ \end{aligned}$

Since water is always present in excess, its relative amount is very high as compared to other species. The relative amount of water is the highest among all other species present in solution.

Learn more:

Calculation of volume of gas: Determine how many moles of water produce:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Concentration terms

Keywords: concentration, concentration terms, solutions, molarity, molality, Ba(OH)2, OH-, Ba2+, 1.0*10^-6 M, 2.0*10^-6 M, water, ionic product constant, 10^-14, 5.0*10^-9 M.

Answer from: Quest

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1

0.44 / 0.09 = 4.8

mgco3-5h2o"

Answer from: Quest

U whay do you need with?

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In a 1.0× 10–6 m solution of ba(oh)2(aq) at 25 °c, identify the relative molar amounts of these spec...

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