Question

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

Answer 1

pH = 1.602

Further explanation

Ca(OH)₂+2HCl⇒CaCl₂+2H₂O

mol Ca(OH)₂ :

`\tt 250* 0.1=25~mlmol=0.025~mol`

mol HCl :

`\tt 750* 0.1=75~mlmol=0.075~mol`

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⁺)

`\tt (0.025)/(750+250)=(0.025)/(1~L) =0.025~M`

pH=-log [H⁺]

pH=-log[0.025]

pH=-log[2.5 x 10⁻²]

pH=2-log 2.5=1.602