Question

A 110.0 mL sample of 0.15 M Ca(OH)₂ is titrated with 0.10 M HCl. Determine the pH of the solution after the addition of 290.0 mL HCl.

Answer 1

Criven date as molarity of Ca(OH)2 = 0.15M Volume of Ca(OH) = 1100ml=o.llol Molarity of Hol a o.lom Volume of HCl = 290 me = 0.2002 Solution How we first Count number of molex of onion & Htion moles Cacon), = 0.15 mx 0illOL = 0.0165 mol COOH, 2 Ca2+ + 204 moles ont = 2X000165 = 0.033 mol moles Ht added = 0.10% 0.2% = 0.0290 since Here mole on is excess than moleft mole OH in excess & 0.033-0.029 mol = 0.00 4 mal total volume of solution - 0.110L + 0.990L = 0.400 L . [or ] = 0.004 mal 0.400L [OH-] = 0.01M

"рон = - Тез Сон"] оя[o • o/] | poн – | PH= 14-pon рн - 14 - 2 pH = 12