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 :

$\tt 250\times 0.1=25~mlmol=0.025~mol$

mol HCl :

$\tt 750\times 0.1=75~mlmol=0.075~mol$

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⁺)

$\tt \dfrac{0.025}{750+250}=\dfrac{0.025}{1~L} =0.025~M$

pH=-log [H⁺]

pH=-log[0.025]

pH=-log[2.5 x 10⁻²]

pH=2-log 2.5=1.602