Solveeit Logo
Login

Question

Question: After 20mL of 0.1M \(Ba{{(OH)}_{2}}\) is mixed with 10mL of 0.2M \(HCl{{O}_{4}}\) , the concentratio...

After 20mL of 0.1M Ba(OH)2Ba{{(OH)}_{2}} is mixed with 10mL of 0.2M HClO4HCl{{O}_{4}} , the concentration of is :
(A) 2×10−3M2\times {{10}^{-3}}M
(B) 10−3M{{10}^{-3}}M
(C) 0.066M0.066M
(D) 0.2M0.2M

Explanation

Solution

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 × molarity of base × n factormilliequivalent\,of{{\,}^{-}}OH\,=\,volume\,of\,base\,\times \,molarity\,of\,base\,\times \,n\,factor
Where n is the number of −OH^{-}OHions furnished by Ba(OH)2Ba{{(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)2Ba{{(OH)}_{2}}= 20mL
Volume of HClO4HCl{{O}_{4}} = 10mL
Concentration of Ba(OH)2Ba{{(OH)}_{2}}= 0.1M
Concentration of HClO4HCl{{O}_{4}}= 0.2M
n = 2
First, we will calculate the mili equivalent of
milliequivalent of −OH = volume of base × molarity of base × n factormilliequivalent\,of{{\,}^{-}}OH\,=\,volume\,of\,base\,\times \,molarity\,of\,base\,\times \,n\,factor
milliequivalent of −OH = 20×0.1 ×2 = 4 mEqmilliequivalent\,of{{\,}^{-}}OH\,=\,20\times 0.1\,\times 2\,=\,4\,mEq
Now, we will calculate milliequivalents of
milliequivalent of HClO4 = vol of acid × molarity of acid × n factormilliequivalent\,of\,HCl{{O}_{4}}\,=\,vol\,of\,acid\,\times \,molarity\,of\,acid\,\times \,n\,factor
milliequivalent of HClO4 = 10×0.2 ×1 = 2 mEqmilliequivalent\,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)2Ba{{(OH)}_{2}} will neutralize 2 milli equivalent of HClO4HCl{{O}_{4}}.
So, number of milliequivalent of Ba(OH)2Ba{{(OH)}_{2}} left = 4-2 = 2
Hence, the number of −OH^{-}OH left = 2 mEq
Now, to calculate the required concentration of −OH^{-}OHions, we will use volume of the solution
Total volume of the solution = 20+10 = 30mL
Concentration of −OH^{-}OH ions left
= mEq of −OHtotal volume of solution=\,\dfrac{mEq\,of{{\,}^{-}}OH}{total\,volume\,of\,solution}
Required concentration of −OH^{-}OHions = 230 = 0.066 N=\,\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.