What is the molar concentration of all ions present in a solution prepared by mixing the followi

What is the molar concentration of all ions present in a...

Key Concepts

Stoichiometry of Reactions

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It allows for the balancing of chemical equations and calculation of reactant consumption and product formation, which is particularly important in acid-base neutralization reactions.

Ionic Dissociation

In aqueous solutions, strong electrolytes such as common salts, strong acids, and strong bases completely dissociate into their constituent ions. Understanding this dissociation is crucial for determining the molar concentrations of individual ions present in the mixture.

Limiting Reagent Analysis

In reactions where two or more reactants are present, the limiting reagent is the substance that is completely consumed first, limiting the amount of product formed. Identifying the limiting reagent is essential for accurately determining the final concentrations of ions, especially in acid-base neutralizations.

Molarity

Molarity is a measure of concentration defined as the number of moles of solute per liter of solution. It is used to quantify how much of a substance is dissolved in a given volume of solvent, allowing for precise calculations when mixing solutions or carrying out titrations.

Dilution and Volume Additivity

When solutions are mixed, the individual volumes are often considered additive unless specified otherwise. This means that the total volume is the sum of the volumes of the individual solutions, which is essential when calculating the final concentrations of ions after dilution.

Transcript

00:01 Okay, so in the first part of this problem, we are mixing up two solutions of sodium hydroxide.

00:11 That means to find out the final concentration of each ion, as we are mixing up two solutions of the same compound, first you have to calculate the individual number of moles for each solution.

00:33 And then we have to just add them up to find the final number of moles and then you have to divide that final number of moles by final volume and that will give us the final concentration of sodium hydroxide from where we will be able to say the final concentration of the ions present in this solution so let's first write down the number of moles from each sodium hydroxide solution we will get so here we have 42 ml 170 molar so the number of moles for this solution will be 0 .170 molar or mole per liter times the volume which is 42 ml and if we convert it to liter by dividing it by thousand we'll have 0 .042 liter and it will give us the number of moles for this solution which is 0 .00714 mole noh and for the second solution we have the concentration as 0 .4 mole per liter 0 .4 mole per liter and we have the volume as 37 .6 milliliter or 0 .0376 liter and this will give us a molar number of moles of 0 .0 .0 .0 0 .5.

02:33 Now so the total mole of sodium hydroxide will be some of these two.

02:53 Just if we just add this to up it will give us a value of 0 .0.

03:01 0 to 2 mole sodium hydroxide and if we just add up the two volumes total volume here is the sum of the two volumes which are 42 plus 37 .6 milliliter 79 .6 milliter and if we convert it to liter by dividing it by thousand it will give us 0 .0796 liter now to find out the final concentration is simply the final number of moles which is 0 .022 over the final volume 0 .0796 and from here we can find out the final concentration which will be 0 .278 molar so this is the final sodium hydroxide concentration in this solution now sodium hydroxide will be dissociated to give us sodium ion and hydroxide ion now since from one sodium hydroxide we will get one sodium and one hydroxide that means if the sodium hydroxide concentration is 0 .278 molar that means the concentration of sodium ion will be 0 .278 molar and the hydroxide and concentration will be also 0 .278 molar so these are the concentration of each hand present in the final mixture of sodium hydroxide here now in the second part we are mixing 44 m l of 0 .1 molar sodium sulfate with 25 militer of 0 .15 molar kcel so what we have to do is we have to find the final concentration of each compound here and from there will will be able to calculate the final concentration of each and present so first you find out the final concentration of the n a 2s of 4 so for an 82s of 4 the initial concentration our m initial is 0 .1 molar and initial volume or v initial is 44 milliliter and if you convert it to liter okay we don't need to convert this to later right now we can keep it like this and so we what we'll have to use is we'll have m initial v initial equal to m final m final v final we will use this formula here so the v final is the is the total volume which is the volume of this nate 2 of 4 plus the volume of the kcel we are adding that is 25 ml so the total volume will be 69 ml now here we know the m initial v initial and we know the v final we can find the m final or the final concentration of n8 2 s of 4 after the addition of 25 ml kcl so um v final for n82 s of 4 will be m initial which is 0.

08:58 1 times the v initial which is 44 over the v final which is 69 will give oh okay so this is not v final this is m final the final concentration of an atres of 4 so this will give us the final concentration which is 0 .0638 molar okay so this is this is 69 liter liter and here we have molar okay so this is the final concentration of the n a 2 of 4 now we have to see how n a 2 s of 4 will be will be dissociated it will be dissociated as n a plus plus so 4 2 minus now if you want to balance it we have 2 sodium here that means we will have 2 sodium here now if the initial concentration of an atresophore is of the final concentration of anitres over is 0 .0638 which you just have calculated then the final concentration of sodium ion will be two times this concentration that means 2 times 0 .0638 molar which will give us 0 .1 to 8 molar and the sulfur down concentration will be like this because from one molar anodeuase before we have one molar sulphate produced that means the sulphate ion concentration will be same as the sodium sulphate concentration now we have to find out the concert of the ions present in kcel so we just utilize the same process for kcel um initial concentration is 0 .15 initial volume is 25 and final volume is volume of kcl plus the volume of sodium sulphate which is 44 and this will give us a total of 69 ml now we can calculate the final concentration of kcl which will be just multiplication of initial concentration times the initial volume 25 ml over the total volume which is 69 ml this will give us a value of 0 .0543 molar so this is the final concentration of kcl i have to see how kcl will be dissociated kcl will be dissociated to give us k plus and and cl minus that means from one molar kcl will have one molar k plus and one molar cl minus see if the kcel concentration is 0 .0543 that means it will give us 0 .0543 molar k plus and 0 .0 543 molar cl plus and 0 .0543 molar cl minus so the concentration of each ion present is k plus a on concentration is 0 .0543, cl minus on concentration is 0 .0543, and the na plus on concentration is 0 .1 to 8, and sulfate and concentration is 0 .0638.

13:52 These are the concentration of each one present in this mixture.

13:58 Now in the final one, we are mixing 3 .6 gram kcl in 75 millil of 0 .25 molar are c .l 2...

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