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Home / Doubts and Solutions/JEE-BASIC/Main/Chemistry

Ques 22

Asked by chandanbr6004 12th December 2017, 11:25 PM
Answered by Expert
Answer:
begin mathsize 16px style We space have space given comma 50 ml space of space straight H subscript 2 straight O space is space added space to space 50 ml space of space 1 cross times 10 to the power of negative 3 space end exponent straight M space Ba left parenthesis OH right parenthesis subscript 2 space solution comma space straight i. straight e. comma space the space volume space of space solution space is space made space double space by space the space addtion space of space 50 space ml space water. Therefore comma space first space we space find space the space molarity space of space Ba left parenthesis OH right parenthesis subscript 2 space solution comma straight M space equals fraction numerator 0.001 over denominator 2 end fraction equals space 5 space cross times 10 to the power of negative 4 end exponent straight M space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space Ba left parenthesis OH right parenthesis subscript 2 space end subscript rightwards arrow with straight H subscript 2 straight O on top space Ba to the power of 2 plus end exponent space plus space 2 space OH to the power of minus conc. space in space moles space space space space space space space space 5 space cross times 10 to the power of negative 4 end exponent straight M space space space space space space space space space space space space space space space space space space space space space space space 2 left parenthesis space 5 space cross times 10 to the power of negative 4 end exponent straight M right parenthesis space space space left square bracket OH right square bracket to the power of minus space equals space 2 left parenthesis space 5 space cross times 10 to the power of negative 4 end exponent straight M right parenthesis space equals space 0.001 space straight M pOH space equals space minus log left parenthesis OH to the power of minus right parenthesis space equals space minus log left parenthesis 0.001 right parenthesis space equals space 3 space therefore space pH space equals space 14 minus pOH space space space space space space space space space space space equals space 14 space minus space 3 space space space space space space space space space space space space equals 11 end style
Answered by Expert 13th December 2017, 11:53 AM
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