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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) $0.2M$

Last updated date: 01st Mar 2024
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IVSAT 2024
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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
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.