Arrange the following resultant mixtures in increasing order of...

For mixture A:
Millimoles of OH$^{-}=10 \times 0.2 \times 2 = 4$
Millimoles of H$^{+}=25 \times 0.1 \times 1 = 2.5$
Remaining OH$^{-}=4 - 2.5 = 1.5$ mmol
The solution is basic, so pH $> 7$.

For mixture B:
Millimoles of H$^{+}=10 \times 0.01 \times 2 = 0.2$
Millimoles of OH$^{-}=10 \times 0.01 \times 2 = 0.2$
The solution is completely neutralized, so pH $= 7$.

For mixture C:
Millimoles of H$^{+}=10 \times 0.1 \times 2 = 2$
Millimoles of OH$^{-}=10 \times 0.1 \times 1 = 1$
Remaining H$^{+}=2 - 1 = 1$ mmol
The solution is acidic, so pH $< 7$.

Therefore, the increasing order of pH values is C $<$ B $<$ A.

Answer: $C < B < A$