Arrange the following resultant mixtures in increasing order of their pH values
A. 10 mL 0.2 M Ca(OH)$$_2$$ + 25 mL 0.1 M HCl
B. 10 mL 0.01 M H$$_2$$SO$$_4$$ + 10 mL 0.01 M Ca(OH)$$_2$$
C. 10 mL 0.1 M H$$_2$$SO$$_4$$ + 10 mL 0.1 M KOH
Choose the correct answer from the options given below:

By Comparing excess $$H^+$$ and $$OH^-$$ in each mixture we can determine whether the solution is Acidic, Basic, or Neutral.

$$A.\ 10ml\ 0.2M\ Ca\left(OH\right)_2\ +\ 25ml\ 0.1M\ HCl$$

$$Ca\left(OH\right)_2\longrightarrow\ 2OH^-$$

Moles of $$OH^-$$:

$$0.2\times\ 0.01\times\ 2=0.004$$

Moles of $$H^+$$:

$$0.1\times\ 0.025=0.0025$$

Excess $$OH^-$$:

$$0.004-0.0025=0.0015$$

Solution is Basic.

$$B.\ 10\ ml\ 0.01\ M\ H_2SO_4+10\ ml\ 0.01M\ Ca\left(OH\right)_2$$

For $$H_2SO_4:$$

$$H_2SO_4\longrightarrow\ 2H^+$$

Moles of $$H^+$$: 

$$0.01\times\ 0.01\times\ 2=2\times\ 10^{-4}$$

Moles of $$OH^-$$: 

$$0.01\times\ 0.01\times\ 2=2\times\ 10^{-4}$$

Exactly Neutral

$$pH\approx\ 7$$

$$C.\ 10mL\ 0.1\ M\ H_2SO_4+10\ mL\ 0.1\ M\ KOH$$

Moles of $$H^+$$:

$$0.1\times\ 0.01\times\ 2=0.002$$

Moles of $$OH^-$$:

$$0.1\times\ 0.01=0.001$$

Solution is acidic.

Therefore,

C$$\longrightarrow\ $$ acidic(lowest pH)

B$$\longrightarrow\ $$ neutral

A$$\longrightarrow\ $$ basic (highest pH)

Therefore, the order will be $$C<B<A$$

i.e Option C