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Chapter 1: Problem 26

The volume (in \(\mathrm{mL}\) ) of \(0.1 \mathrm{~N} \mathrm{NaOH}\) required to neutralise \(10 \mathrm{~mL}\) of 0.1 \(\mathrm{N}\) phosphinic acid is $$ \text { [Main Sep. } \mathbf{0 3}, \mathbf{2 0 2 0}[\mathbf{I I})] $$

Short Answer

Expert verified
20 mL of 0.1 N NaOH is required.

Step by step solution

01

Understanding the Problem

The problem asks for the volume of a sodium hydroxide solution required to neutralize a phosphinic acid solution. Both solutions have a concentration of 0.1N.
02

Write the Neutralization Reaction

The chemical reaction for neutralizing phosphinic acid (also known as hypophosphorous acid, H鈧働O鈧�) with sodium hydroxide (NaOH) can be written as: \[\text{H}_3\text{PO}_2 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{HPO}_2 + 2\text{H}_2\text{O}\].This shows that 2 moles of NaOH are required to neutralize 1 mole of phosphinic acid.
03

Determine Equivalent Milliequivalents

The equivalent milliequivalents (meq) of H鈧働O鈧� is given by: \[\text{meq of H}_3\text{PO}_2 = 10 \text{ mL} \times 0.1 \text{ N} = 1\]meq because phosphinic acid donates 1 acidic hydrogen to the reaction.
04

Calculate Required meq of NaOH

Since 2 meq of NaOH are needed to neutralize 1 meq of H鈧働O鈧�:\[\text{meq of NaOH required} = 2 \times 1 = 2\].
05

Calculate Volume of NaOH Solution

Using the formula: \[\text{Volume (mL)} = \frac{\text{meq}}{\text{Normality}}\]we find the volume of NaOH needed:\[\text{Volume of NaOH} = \frac{2 \text{ meq}}{0.1 \text{ N}} = 20 \text{ mL}\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Stoichiometry
In chemistry, stoichiometry is the method used to determine the quantities of reactants and products in chemical reactions.
It involves the calculation of mass, volume, and concentrations based on the balanced chemical equations of reactions. Calculating stoichiometric ratios requires understanding how many moles of each substance are involved.
In this exercise, stoichiometry tells us how sodium hydroxide (NaOH) and phosphinic acid (H鈧働O鈧�) interact during the neutralization.
  • The balanced chemical equation \[ ext{H}_3 ext{PO}_2 + 2 ext{NaOH} ightarrow ext{Na}_2 ext{HPO}_2 + 2 ext{H}_2 ext{O} \] explains that two moles of NaOH are required to neutralize each mole of phosphinic acid.
  • This ratio is crucial for calculating the exact amounts needed in the reaction.
  • Thus, stoichiometry helps in ensuring that the reaction proceeds completely without excess of any reactant.
Exploring Molarity and Normality
Molarity is a measure of the concentration of a solute in a solution, expressed as moles per liter.
In this solution, it is essential to recognize the relationship between molarity and normality, especially while working with acids and bases. For phosphinic acid,
  • The concentration provided (\(0.1 ext{ N}\)) indicates that it donates one acidic hydrogen, aligning with its role in the reaction.
  • Similarly, NaOH, also with a normality of \(0.1 ext{ N}\), reflects its strength as a base in this reaction.
To calculate the volume needed for neutralization, normality simplifies since normality directly correlates with the equivalent moles needed.
Instead of calculating molarity separately, normality already considers the ion exchange specific to acid-base reactions. This simplicity aids in solving problems quickly, as it bridges the specific reaction behavior directly to the calculated values used in the equations.
Balancing Chemical Equations
Chemical equations represent the symbolic depiction of chemical reactions where reactants transform into products.
They need to be balanced to reflect the conservation of mass, where atom counts for each element are equal on both sides of the equation. In our original equation with phosphinic acid and NaOH,
  • Each element's atom count is crucial; therefore, understanding that two NaOH molecules react with one phosphinic acid is determined from balancing.
  • It ensures that no atoms are lost or gained, highlighting the fundamental law that mass is conserved in reactions.
Balancing chemical equations helps in calculating the exact amount of each substance required.
In a real-world lab scenario, ensuring equations are balanced is essential for your calculated reactants to result in desired products without wastage, excess, or shortage.

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Most popular questions from this chapter

Which has maximum number of atoms? (a) \(24 g\) of \(C(12)\) (b) \(56 \mathrm{~g}\) of \(\mathrm{Fe}(56)\) (c) \(27 \mathrm{~g}\) of \(\mathrm{Al}(27)\) (d) \(108 \mathrm{~g}\) of \(\operatorname{Ag}(108)\)

Mixture \(X=0.02 \mathrm{~mol}\) of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{Br}\) and \(0.02 \mathrm{~mol} \mathrm{o}\) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right] \mathrm{SO}_{4}\) was prepared in 2 litre of solution. [2003S] 1 litre of mixture \(X+\) excess \(\mathrm{AgNO}_{3} \rightarrow Y\) 1 litre of mixture \(X+\) excess \(\mathrm{BaCl}_{2} \rightarrow Z\) No. of moles of \(Y\) and \(Z\) are (a) \(0.01,0.01\) (b) \(0.02,0.01\) (c) \(0.01,0.02\) (d) \(0.02,0.02\)

1 gram of a carbonate \(\left(\mathrm{M}_{2} \mathrm{CO}_{3}\right)\) on treatment with excess \(\mathrm{HCl}\) produces \(0.01186\) mole of \(\mathrm{CO}_{2}\). The molar mass of \(\mathrm{M}_{2} \mathrm{CO}_{3}\) in \(\mathrm{g} \mathrm{mol}^{-1}\) is : [Main 2017] (a) 1186 (b) \(84.3\) (c) \(118.6\) (d) \(11.86\)

(a) One litre of a sample of hard water contains \(1 \mathrm{mg}\) of \(\mathrm{CaCl}_{2}\) and 1 \(\mathrm{mg}\) of \(\mathrm{MgCl}_{2}\). Find the total hardness in terms of parts of \(\mathrm{CaCO}_{3}\) per \(10^{6}\) parts of water by weight. (b) A sample of hard water contains \(20 \mathrm{mg}\) of \(\mathrm{Ca}^{++}\)ions per litre. How many milli-equivalent of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) would be required to soften 1 litre of the sample? (c) \(1 \mathrm{~g}\) of \(\mathrm{Mg}\) is burnt in a closed vessel which contains \(0.5 \mathrm{~g}\) of \(\mathrm{O}_{2}\). (i) Which reactant is left in excess? (ii) Find the weight of the excess reactants? (iii) How may milliliters of \(0.5 \mathrm{~N} \mathrm{H}_{2} \mathrm{SO}_{4}\) will dissolve the residue in the vessel.

A \(2.0 \mathrm{~g}\) sample of a mixture containing sodium carbonate, sodium bicarbonate and sodium sulphate is gently heated till the evolution of \(\mathrm{CO}_{2}\) ceases. The volume of \(\mathrm{CO}_{2}\) at \(750 \mathrm{~mm}\) Hg pressure and at \(298 \mathrm{~K}\) is measured to be \(123.9 \mathrm{~mL}\). A \(1.5 \mathrm{~g}\) of the same sample requires 150 mL. of \((\mathrm{M} / 10) \mathrm{HCl}\) for complete neutralisation. Calculate the \(\%\) composition of the components of the mixture. $$ \text { Page } 36,1,773-Q+\quad[1992-5 \text { Marks }] $$
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