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What is the [OH-] in the final solution prepared by mixing 20 mL of 0.050 M HCl with 30 mL of 0.10 M Ba(OH)2 ?
Asked by kumarisakshi0209 | 22 Mar, 2019, 11:41: AM
answered-by-expert Expert Answer
Given:
 
Volume of HCl = 20 ml
 
Molarity = 0.050 M
 
Volume of Ba(OH)2 = 30 ml
 
Molarity of Ba(OH)2 = 0.1 M
 
No. of milliequivalents of HCl = 2× 0.05 × 1
 
                                           =1
 
No. of milliequivalents of Ba(OH)2 = 2× 30×0.1
 
                                                  =6
 
[OH−] of solution 
 
space equals space fraction numerator milliequivalents space of space Ba left parenthesis OH right parenthesis subscript 2 space minus space milliequivalents space of space HCl space over denominator Total space volume space of space solution end fraction equals space fraction numerator 6 minus 1 over denominator 30 plus 20 end fraction equals 5 over 50 equals space 0.1 space straight M
 
[OH-] in the final solution 0.1 M.
 
Answered by Varsha | 22 Mar, 2019, 07:06: PM

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