Part A A 250 ml sample of 0.10 M Ca(OH)2 is titrated with 0.10 M HCI. Determine the pH of the solution after the addition of 750 ml HCI

pH = 1.602

Ca(OH)₂+2HCl⇒CaCl₂+2H₂O

mol Ca(OH)₂ :

mol HCl :

ICE method :

              Ca(OH)₂+2HCl⇒CaCl₂+2H₂O

initial        0.025    0.075

change     0.025   0.05     0.025  0.05

equilbrium   0       0.025   0.025  0.05

There is a residual strong acid, then the pH is calculated from the acid concentration (H⁺)

pH=-log [H⁺]

pH=-log[0.025]

pH=-log[2.5 x 10⁻²]

pH=2-log 2.5=1.602

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i do not get it   at all