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1 L of 0.1 M NaOH, 1 L of 0.2 M KOH, and...

`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.

A

0.01 M

B

0.01 N

C

0.1 N

D

0.001 M

Text Solution

Verified by Experts

Total volume `(V_4)=4L`
n factor for NaOH and KOH is 1 whereas that for `Ba(OH)_2` is 2
Now, `N_1V_1+N_2V_2+N_3V_3=N_4V_4`
`0.1xx1+0.2xx1+0.05xx2xx2=N_4xx4`
`rArr 0.1 + 0.2 + 0.2 = N_4xx4 rArr N_4 = 0.125`
Hence, final concentration of solution = 0.1 N
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