Calculate the $pH$ of the resultant mixtures:
a) 10 mL of 0.2M ${Ca\left(OH\right)}_2$ + 25 mL of 0.1M $HCl$
b) 10 mL of 0.01M ${H}_2{SO}_4$+ 10 mL of 0.01M ${Ca\left(OH\right)}_2$
c) 10 mL of 0.1M ${H}_2{SO}_4$ + 10 mL of 0.1M $KOH$

(a) Total number of moles present in 10 mL of 0.2 M calcium hydroxide are $\frac{10\times 0.2}{1000}=0.002$ moles.
Total number of moles present in 25 mL of 0.1 M HCl are $\frac{25\times 2}{1000}=0.0025$ moles.
$Ca(OH{)}_2+2HCl\to CaC{l}_2+2H_{2}O$
1 mole of calcium hydroxide reacts with 2 moels  of HCl.
0.0025 moles of HCl will react with 0.00125 moles of calium hydroxide.
Total number of moles of calcium hydroxide unreacted are $0.002-0.00125=0.00075$ moles.
Total volume of the solution is $10+25=35$ mL.
The molarity of the solution is $\frac{0.00075\times 1000}{35}=0.0214M$
$[O{H}^{-}]=2\times 0.0214=0.0428M$
$pOH=-log0.0428=1.368$
$pH=14-pOH=14-1.368=12.635$

(b) Total number of moles present in 10 mL of 0.01 M $H_{2}S{O}_4$ is $\frac{10\times 0.01}{1000}=0.0001$ mol.
Total number of moles present in 10 mL of 0.01 M $Ca(OH{)}_2=\frac{10\times 0.01}{1000}=0.0001$ mole.
$Ca(OH{)}_2+H_{2}S{O}_4\to CaS{O}_4+2H_{2}O$
$0.0001$ mole of $Ca(OH{)}_2$ will react completely with 0.0001 mole of $H_{2}S{O}_4$.
Hence, the resuting solution is neutral with pH 7.0

(c) Total number of moles in 10 mL of 0.1 M $H_{2}S{O}_4=\frac{10\times 0.1}{1000}=0.001$ mole.
Total number of moles present in 10 mL of 0.1 M KOH $=\frac{10\times 0.1}{1000}=0.001$ mole.
$2KOH+H_{2}S{O}_4\to K_{2}S{O}_4+2H_{2}O$
2 moles of KOH reacts with 1 mole of $H_{2}S{O}_4$.
0.001 mole of KOH will react with 0.0005 mole of $H_{2}S{O}_4$.
Number of moles of $H_{2}S{O}_4$ left $=0.001-0.0005=0.005M$
Volume of solution is $10+10=20$mL
Molarity of the solution is $\frac{0.0005\times 1000}{20}=0.025M$.
$[H^{+}]=2\times 2.5\times 10^{-2}=0.05M$
$pH=-log[H^{+}]=-log0.05=1.3$.