Calculate the pH of the solution obtained by mixing 100 mL 0.075(M) `Ca(OH)_2` solution with 100mL 0.1(M) HCl solution

Explanation

100mL of 0.1(M) HCl`=(0.1)/(1000)xx100-=0.01` mol of
`H^(+)` ions
100mL 0.1(M) `Ca(OH)_2=2xx(0.1)/(1000)xx100`
`=0.02mol OH^-`
Total volume`=(100+100)mL=200mL`
In mixed solution, remaining `OH^-`=0.2-0.1
=0.01 mol
Let the molarity of `H_2SO_4` solution be x (M)
`100mLx(M)H_2SO_4-=2xx(x)/(1000)xx100=0.2x` mol
`H^+` ions
`:.0.2x=0.01:.x=0.05`
`:.` Molarity of `H_2SO_4` solution=0.05(M)
`:.pH=-log_(10)H^+=-log_(10)(2xx0.05)=1`