A 100.0 ml sample of 0.10 m ca (oh) 2 is titrated with 0.10 m hbr. determine the ph of the solution after the addition of 400.0 ml hbr. the chemical equation is below. ca (oh) 2 (aq) + 2hbr (aq) → cabr2 (aq) + h20 (l)

The balanced equation for the reaction is as follows;

Ca (OH) ₂ + 2HBr - - > CaBr₂ + 2H₂O

stoichiometry of Ca (OH) ₂ to HBr is 1:2

number of Ca (OH) ₂ moles reacted - 0.10 mol/L x 0.1000 L = 0.010 mol

Number of HBr moles added - 0.10 mol/L x 0.4000 = 0.040 mol

1 mol of Ca (OH) ₂ needs 2 mol of HBr for neutralisation

therefore 0.010 mol of Ca (OH) ₂ needs - 0.010 x 2 = 0.020 mol of HBr to be neutralised

but 0.040 mol of HBr has been added therefore number of moles of HBr in excess - 0.040 - 0.020 = 0.020 mol

then pH of the medium can be calculated using the excess H⁺ ions

HBr is a strong acid therefore complete ionization

[HBr] = [H⁺]

[H⁺] = 0.020 mol / (100.0 + 400.0 mL)

= 0.020 mol / 0.5 L

= 0.040 mol/L

pH = - log[H⁺]

pH = - log [0.040 M]

pH = 1.40

pH of the medium is 1.40