When \[\mathbf{100}~ml~\mathbf{of}~N/\mathbf{10}~NaOH~\] are added to \[\mathbf{50}~\mathbf{ml}~\mathbf{of}~\mathbf{N}/\mathbf{5}~\mathbf{HCl}\], the pH of the resulting solution is:
A. 7
B. greater than 7
C. less than 7
D. zero
Answer
630k+ views
Hint: pH of the solution defined as the measure of the molar concentration of Hydrogen ions present in the solution or also measure of the acidity or alkalinity of solution. If we know the molarity of an acid or base solution, then we can easily convert it to normality.
Complete step by step answer:
As we know, the range of scale varies from 0 to 14. less than 7 are termed as acidic while greater than 7 are basic or alkaline in nature and at it is neutral.
Now, let us calculate the of the resulting solution of \[NaOH\]
As we know,
\[Equivalents\text{ }of~NaOH~=~Normality\text{ }of~NaOH~\times ~Volume\text{ }of~NaOH\]
\[\begin{align}
& \Rightarrow 1/10\times 100 \\
& \Rightarrow 10milliequivalents \\
\end{align}\]
Similarly, for \[HCl\]
\[Equivalents\text{ }of~HCl~=~Normality\text{ }of~HCl~\times ~Volume\text{ }of~HCl\]
\[\begin{align}
& \Rightarrow 1/5\times 50 \\
& \Rightarrow 10milliequivalents \\
\end{align}\]
As we can see from above equation,
\[Here,\text{ }equivalents\text{ }of~NaOH~=~equivalents\text{ }of~HCl\]
Therefore, the resulting solution is 7.
From the above equation, we can conclude that the resulting solution is neutral because it has 7 and hence the correct option is A.
Note: Remember don’t confuse normality with molarity. There is a very close relation between molarity and normality. Normality (N) which is defined as the number of equivalents of solute which is dissolved per litre of solution. Whereas, molarity refers to the concentration of compound or ion in a solution.
Mathematically it is represented as,
\[\left( equivalents/Volume(L)=N \right)\]
Whereas, \[N=\text{ }Normality\]
\[L=Litre\]
Molarity can’t be used instead of normality because of the law of equivalence.
Complete step by step answer:
As we know, the range of scale varies from 0 to 14. less than 7 are termed as acidic while greater than 7 are basic or alkaline in nature and at it is neutral.
Now, let us calculate the of the resulting solution of \[NaOH\]
As we know,
\[Equivalents\text{ }of~NaOH~=~Normality\text{ }of~NaOH~\times ~Volume\text{ }of~NaOH\]
\[\begin{align}
& \Rightarrow 1/10\times 100 \\
& \Rightarrow 10milliequivalents \\
\end{align}\]
Similarly, for \[HCl\]
\[Equivalents\text{ }of~HCl~=~Normality\text{ }of~HCl~\times ~Volume\text{ }of~HCl\]
\[\begin{align}
& \Rightarrow 1/5\times 50 \\
& \Rightarrow 10milliequivalents \\
\end{align}\]
As we can see from above equation,
\[Here,\text{ }equivalents\text{ }of~NaOH~=~equivalents\text{ }of~HCl\]
Therefore, the resulting solution is 7.
From the above equation, we can conclude that the resulting solution is neutral because it has 7 and hence the correct option is A.
Note: Remember don’t confuse normality with molarity. There is a very close relation between molarity and normality. Normality (N) which is defined as the number of equivalents of solute which is dissolved per litre of solution. Whereas, molarity refers to the concentration of compound or ion in a solution.
Mathematically it is represented as,
\[\left( equivalents/Volume(L)=N \right)\]
Whereas, \[N=\text{ }Normality\]
\[L=Litre\]
Molarity can’t be used instead of normality because of the law of equivalence.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Trending doubts
How many of the following diseases can be studied with class 11 biology CBSE

Which of the following enzymes is used for carboxylation class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Receptor sites for neurotransmitters are present on class 11 biology CBSE

JGA JuxtaGlomerular Apparatus a sensitive region which class 11 biology CBSE

Root hairs develop from the region of A Meristematic class 11 biology CBSE

