Question

`1 L` of `0.1 M NaOH, 1 L` of `0.2 M KOH`, and `2 L` of `0.05 M Ba (OH)_(2)` are mixed togther. What is the final concentration of the solution.

Answer 1

Total volume `(V_(4)) = 1 + 1 + 2 = 4 L`
`n` factor of `NaOH KOH` is 1, while for `Ba(OH)_(2)` is 2.
`N_(1) V_(1) + N_(2) V_(2) + N_(3) V_(3) = N_(4) V_(4)`
`0.1 xx 1 + 0.2 xx 1 + 0.05 xx 2 xx 2 = N_(4) xx 4`
`0.1 + 0.2 + 0.2 = N_(4) xx 4`
`N_(4) = (0.4)/(4) = 0.1 N`
Hence, final concentration of solution `= 0.1 N`