Text Solution
Solution : Since hydrochloric acid is 0.2 M , therefore the number of gram moles of HCl in 1000 mL of the solution =0.2 <br> The number of gram moles of HCl in 150 mL of the solution <br> `=(0.2xx150)/1000=30/1000` <br> Similarly , the number of gm moles of NaOH present in 150 mL of 0.1 M solution <br> `=(0.1xx150)/1000=15/1000` <br> Since large number of gm moles of HCl are present in solution, `15/1000` gm moles of HCl will neutralise `15/1000` gm moles of NaOH on mixing the two solutions. <br> The number of moles of HCl left after mixing the two solutions =`30/1000-15/1000=15/1000` <br> Total volume of the solution after mixing =150+150=300 mL <br> Now, 300 mL of solution contain HCl = `15/1000` gm mol <br> `therefore` 1000 mL of solution contain HCl `=15/1000xx1/300xx1000` <br> `=1/20` gm mol per litre <br> Molar concentration of `H_3O^+` i.e., `[H_3O^+]=1/20 "mol L"^(-1)=0.05 "mol L"^(-1)` <br> `therefore` pH of solution =-log [0.05]=1.301
100 ml of 1 M NaOH is mixed with 50 ml of1N KOH solution. What is Normality of mixture?
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Calculate molarity of the resulting mixture, when 0.1 N of 100 ml H2SO4 is mixed with 0.2 N of 100 ml of NaOH solution:
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