`100 mL` of `0.1 M HCl + 100 mL` of `0.2 M H_(2)SO_(4) + 100 mL` of `0.1 M HNO_(3)` are mixed togther. a. What is the final conecntration of the solution. b. What would be the final concentration of the solution. If the solution is made to `1 L` by adding `H_(2) O`? c. What would be the final concentration of the solution if `700 mL` of `H_(2)O` is added to the solution?

`100 mL` of `0.1 M HCl + 100 mL` of `0.2 M H_(2)SO_(4) + 100 mL` of `0

1.	`100 mL` of `0.1 M HCl + 100 mL` of `0.2 M H_(2)SO_(4) + 100 mL` of `0.1 M HNO_(3)` are mixed togther. a. What is the final conecntration of the solution. b. What would be the final concentration of the solution. If the solution is made to `1 L` by adding `H_(2) O`? c. What would be the final concentration of the solution if `700 mL` of `H_(2)O` is added to the solution?
Answer» a. `N_(1) V_(1) + N_(2) V_(2) + N_(3) V_(3) = N_(4) V_(4)` `(V_(4) = V_(1) + V_(2) V_(3) = 100 + 100 + 100 = 300 mL)` `n` factor for `HCl` and `HNO_(3) = 1` `n` factor for `H_(2) SO_(4) = 2` In such questions always convert molarity to normality, since 1 grams equivalent of one reactant is always equal to 1 gram equivalent of another reactant. `100 xx 0.1 + 100 xx 0.2 + 2 100 xx 0.1 = N_(4) xx 300 + 10 + 40 + 10 = N_(4) xx 300` `:. N_(4) = (60)/(300) = (1)/(5) = 0.2 N` Hence, final concentration of solution `= 0.2 N` b. Final volume `= 1 L = 1000 mL` `:. 100 xx 0.1 + 100 xx 0.2 xx 2 + 100 xx 0.1 = N_(4) xx 1000` `N_(4) = (60)/(1000) = 0.06 N` Hence, final concentration of solution `= 0.06 N` c. Final volume `= 300 + 700 = 1000 mL` `:. N_(4) = (60)/(1000) = 0.06` Hence, final concentration of solution `= 0.06 N`

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