Question

calculate the normality if HCl which contains 2.281 g of the acid in 200ml . find out the volume of this solution required to nuitralize exactly 50 ml of 0.12N Naoh solution.
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Answer 1

Answer:

3.00∙mol12∙mol∙L−1×1000∙mL∙L−1=250∙mL

Explanation:

Well, how do we define molar concentration" role="presentation" style="margin: 0px; padding: 0px; display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">molar concentrationmolar concentration?

molar concentration=Moles of soluteVolume of solution" role="presentation" style="margin: 0px; padding: 0px; display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">molar concentration=Moles of soluteVolume of solutionmolar concentration=Moles of soluteVolume of solution

And so we get (i) we get units of mol•L−1" role="presentation" style="margin: 0px; padding: 0px; display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">mol∙L−1mol•L−1, and (ii) the PRODUCT ….

Volume of solution×molar concentration=Moles of solute" role="presentation" style="margin: 0px; padding: 0px; display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">Volume of solution×molar concentration=Moles of soluteVolume of solution×molar concentration=Moles of solute

And in the given scenario …. to get the volume of acid that contains 3.00•mol" role="presentation" style="margin: 0px; padding: 0px; display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">3.00∙mol3.00•mol HCl(aq)" role="presentation" style="margin: 0px; padding: 0px; display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">HCl(aq)HCl(aq), we take the quotient…

3.00•mol12•mol•L−1×1000•mL•L−1=250•mL" role="presentation" style="margin: 0px; padding: 0px; display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">3.00∙mol12∙mol∙L−1×1000∙mL∙L−1=250∙mL3.00•mol12•mol•L−1×1000•mL•L−1=250•mL..

Just to add that the question was NOT set by a chemist. The concentrated hydrochloric acid we use in the lab is 10.6•mol•L−1" role="presentation" style="margin: 0px; padding: 0px; display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; position: relative;">10.6∙mol∙L−110.6•mol•L−1, and higher concentrations are unavailable…

Answer 2

Answer: The normality of the solution is 0.3125 and 19.2 ml of this solution is required to neutralize the NaOH solution

Explanation:

1 molecule of HCl ionizes to give 1 mole of H⁺ and 1 molecule of Cl⁻

HCl → H⁺  +  Cl⁻

∴ n-factor of HCl is 1

Amount of HCl contained = 2.281 g

Molar mass of HCl = 36.45 g

Number of moles of HCl = Amount taken ÷ Molar mass

Number of moles of HCl = 2.281 ÷ 36.45

Number of moles of HCl = 0.0625

Volume of solution = 200 ml

Volume of solution = 0.2 litres

Molarity of solution  =  Number of moles ÷ Volume of solution

Molarity of solution =  0.0625 ÷ 0.2

Molarity of solution = 0.3125

Normality = Molarity × n-factor

Normality = 0.3125 × 1

Normality = 0.3125

Let the required volume of this solution be V ml.

Equal equivalents of HCl solution and NaOH solution will react with each other.

⇒ Equivalents of HCl solution =   Equivalents of NaOH solution

⇒ Normality × volume of HCl =  Normality × volume of NaOH

⇒ 0.3125 × V  =  0.12 × 50

⇒ 0.3125V  =  6

⇒ $V = \frac{6}{0.3125}$

⇒ V = 19.2

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