100 mL 0.1(M) HCl solution is added to 100mL 0.1(M) `Ca(OH)_2` solution. The excess `Ca(OH)_2` remaining in the resulting solution is completely neutalised by 100mL `H_2SO_4` solution. Calculate the concentration and pH of the `H_2SO_4` solution.

Explanation

`pH=1/2pK_a-1/2logc`
`=-1/2log_(10)(1.35xx10^(-3))-1/2log0.1=1.93`
`[H_3O^+]=10^(-pH)=10^(-1.93)=0.011(M)`
`H_2O(l)+HA(aq) At equilibrium: `0.1-0.011" "0.011(M)" "0.011(M)`
`:.[A^-]=0.011(M)` and `[HA]=0.089(M)`