
After 20mL of 0.1M $Ba{{(OH)}_{2}}$ is mixed with 10mL of 0.2M $HCl{{O}_{4}}$ , the concentration of is :
(A) $2\times {{10}^{-3}}M$
(B) ${{10}^{-3}}M$
(C) $0.066M$
(D) $0.2M$
Answer
123.3k+ views
Hint:To answer this question we should know that 1 mili equivalent of a base neutralizes 1 mili equivalent of an acid.
The formula to be used in this question is:
$milliequivalent\,of{{\,}^{-}}OH\,=\,volume\,of\,base\,\times \,molarity\,of\,base\,\times \,n\,factor$
Where n is the number of $^{-}OH$ions furnished by $Ba{{(OH)}_{2}}$.
Complete step by step solution:
Let’s look at the solution of the given question:
In the question it is given that
Volume of $Ba{{(OH)}_{2}}$= 20mL
Volume of $HCl{{O}_{4}}$ = 10mL
Concentration of $Ba{{(OH)}_{2}}$= 0.1M
Concentration of $HCl{{O}_{4}}$= 0.2M
n = 2
First, we will calculate the mili equivalent of
$milliequivalent\,of{{\,}^{-}}OH\,=\,volume\,of\,base\,\times \,molarity\,of\,base\,\times \,n\,factor$
$milliequivalent\,of{{\,}^{-}}OH\,=\,20\times 0.1\,\times 2\,=\,4\,mEq$
Now, we will calculate milliequivalents of
$milliequivalent\,of\,HCl{{O}_{4}}\,=\,vol\,of\,acid\,\times \,molarity\,of\,acid\,\times \,n\,factor$
$milliequivalent\,of\,HCl{{O}_{4}}\,=\,10\times 0.2\,\times 1\,=\,2\,mEq$
Now, we can know that 1 mili equivalent of a base neutralizes only 1 mili equivalent of an acid
Therefore, 2 milli equivalent of $Ba{{(OH)}_{2}}$ will neutralize 2 milli equivalent of $HCl{{O}_{4}}$.
So, number of milliequivalent of $Ba{{(OH)}_{2}}$ left = 4-2 = 2
Hence, the number of $^{-}OH$ left = 2 mEq
Now, to calculate the required concentration of $^{-}OH$ions, we will use volume of the solution
Total volume of the solution = 20+10 = 30mL
Concentration of $^{-}OH$ ions left
$=\,\dfrac{mEq\,of{{\,}^{-}}OH}{total\,volume\,of\,solution}$
Required concentration of $^{-}OH$ions $=\,\dfrac{2}{30}\,=\,0.066\,N$
Hence, the answer of the given question is option (C).
Note: Equivalent and mili equivalent are similar representations. The difference is that if the volume is given in mL then we use mili equivalent and if the volume is in L we use only equivalent. ‘n’ factor for a base is the acidity of bases and for acids it is the basicity of acids.
The formula to be used in this question is:
$milliequivalent\,of{{\,}^{-}}OH\,=\,volume\,of\,base\,\times \,molarity\,of\,base\,\times \,n\,factor$
Where n is the number of $^{-}OH$ions furnished by $Ba{{(OH)}_{2}}$.
Complete step by step solution:
Let’s look at the solution of the given question:
In the question it is given that
Volume of $Ba{{(OH)}_{2}}$= 20mL
Volume of $HCl{{O}_{4}}$ = 10mL
Concentration of $Ba{{(OH)}_{2}}$= 0.1M
Concentration of $HCl{{O}_{4}}$= 0.2M
n = 2
First, we will calculate the mili equivalent of
$milliequivalent\,of{{\,}^{-}}OH\,=\,volume\,of\,base\,\times \,molarity\,of\,base\,\times \,n\,factor$
$milliequivalent\,of{{\,}^{-}}OH\,=\,20\times 0.1\,\times 2\,=\,4\,mEq$
Now, we will calculate milliequivalents of
$milliequivalent\,of\,HCl{{O}_{4}}\,=\,vol\,of\,acid\,\times \,molarity\,of\,acid\,\times \,n\,factor$
$milliequivalent\,of\,HCl{{O}_{4}}\,=\,10\times 0.2\,\times 1\,=\,2\,mEq$
Now, we can know that 1 mili equivalent of a base neutralizes only 1 mili equivalent of an acid
Therefore, 2 milli equivalent of $Ba{{(OH)}_{2}}$ will neutralize 2 milli equivalent of $HCl{{O}_{4}}$.
So, number of milliequivalent of $Ba{{(OH)}_{2}}$ left = 4-2 = 2
Hence, the number of $^{-}OH$ left = 2 mEq
Now, to calculate the required concentration of $^{-}OH$ions, we will use volume of the solution
Total volume of the solution = 20+10 = 30mL
Concentration of $^{-}OH$ ions left
$=\,\dfrac{mEq\,of{{\,}^{-}}OH}{total\,volume\,of\,solution}$
Required concentration of $^{-}OH$ions $=\,\dfrac{2}{30}\,=\,0.066\,N$
Hence, the answer of the given question is option (C).
Note: Equivalent and mili equivalent are similar representations. The difference is that if the volume is given in mL then we use mili equivalent and if the volume is in L we use only equivalent. ‘n’ factor for a base is the acidity of bases and for acids it is the basicity of acids.
Recently Updated Pages
The hybridization and shape of NH2 ion are a sp2 and class 11 chemistry JEE_Main

Total number of orbitals associated with the 3rd shell class 11 chemistry JEE_Main

Which of the following has the lowest boiling point class 11 chemistry JEE_Main

Which of the following compounds has zero dipole moment class 11 chemistry JEE_Main

Number of g of oxygen in 322 g Na2SO410H2O is Molwt class 11 chemistry JEE_Main

In the neutralization process of H3PO4 and NaOH the class 11 chemistry JEE_Main

Trending doubts
JEE Mains 2025: Check Important Dates, Syllabus, Exam Pattern, Fee and Updates

JEE Main Login 2045: Step-by-Step Instructions and Details

JEE Main Chemistry Question Paper with Answer Keys and Solutions

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

JEE Main Chemistry Exam Pattern 2025

Other Pages
NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics

NCERT Solutions for Class 11 Chemistry Chapter 8 Organic Chemistry

NCERT Solutions for Class 11 Chemistry Chapter 6 Equilibrium

NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrocarbons

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs
